The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.

In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two. 4.4 The Fundamental Theorem of Calculus Section Project: Demonstrating the Fundamental Theorem 4.5 Integration by Substitution 4.6 Numerical Integration. Logarithmic, Exponential, and Other Transcendental Functions. 5.1 The Natural Logarithmic Function: Differentiation 5.2 The Natural Logarithmic Function: Integration 5.3 Inverse Functions. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. The Area under a Curve and between Two Curves. The area under the graph of the function (f left( x right) ) between the vertical lines (x = a, ) (x = b ) (Figure (2 )) is given by the formula.

Part (1) (FTC1)

If (f) is a continuous function on (left[ {a,b} right],) then the function (g) defined by

[{{gleft( x right)} = intlimits_a^x {fleft( {t} right)dt},;;}kern0pt{{a le x le b}}]

is an antiderivative of (f), that is

[{g^primeleft( x right) = fleft( x right);;text{or};;}kern0pt{frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]

If (f) happens to be a positive function, then (gleft( x right)) can be interpreted as the area under the graph of (f) from (a) to (x.)

The first part of the theorem says that if we first integrate (f) and then differentiate the result, we get back to the original function (f.)

Part (2) (FTC2)

The second part of the fundamental theorem tells us how we can calculate a definite integral.

If (f) is a continuous function on (left[ {a,b} right]) and (F) is an antiderivative of (f,) that is (F^prime = f,) then

[{intlimits_a^b {fleft( x right)dx} }= {Fleft( b right) – Fleft( a right);;}kern0pt{text{or};;{intlimits_a^b {{F^primeleft( x right)}dx} }= {Fleft( b right) – Fleft( a right)}.}]

To evaluate the definite integral of a function (f) from (a) to (b,) we just need to find its antiderivative (F) and compute the difference between the values of the antiderivative at (b) and (a.)

So the second part of the fundamental theorem says that if we take a function (F,) first differentiate it, and then integrate the result, we arrive back at the original function, but in the form (Fleft( b right) – Fleft( a right).)

Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.

The Area under a Curve and between Two Curves

The area under the graph of the function (fleft( x right)) between the vertical lines (x = a,) (x = b) (Figure (2)) is given by the formula

[S = intlimits_a^b {fleft( x right)dx} = {Fleft( b right) – Fleft( a right).}]

Let (Fleft( x right)) and (Gleft( x right)) be antiderivatives of functions (fleft( x right)) and (gleft( x right),) respectively.

If (fleft( x right) ge gleft( x right)) on the closed interval (left[ {a,b} right],) then the area between the curves (y = fleft( x right),) (y = gleft( x right)) and the lines (x = a,) (x = b) (Figure (3)) is given by

[
{S = intlimits_a^b {left[ {fleft( x right) – gleft( x right)} right]dx} }
= {Fleft( b right) – Gleft( b right) }-{ Fleft( a right) + Gleft( a right).}
]

The Method of Substitution for Definite Integrals

The definite integral (intlimits_a^b {fleft( x right)dx} ) of the variable (x) can be changed into an integral with respect to (t) by making the substitution (x = gleft( t right):)

[{intlimits_a^b {fleft( x right)dx} }={ intlimits_c^d {fleft( {gleft( t right)} right)g’left( t right)dt} .}]

The new limits of integration for the variable (t) are given by the formulas

[{c = {g^{ – 1}}left( a right),;;}kern-0.3pt{d = {g^{ – 1}}left( b right),}]

where ({g^{ – 1}}) is the inverse function to (g,) that is (t = {g^{ – 1}}left( x right).)

Integration by Parts for Definite Integrals

In this case the formula for integration by parts looks as follows:

[{intlimits_a^b {udv} }={ left. {uv} right|_a^b – intlimits_a^b {vdu} ,}]

where (left. {uv} right|_a^b) means the difference between the product of functions (uv) at (x = b) and (x = a.)

Solved Problems

Click or tap a problem to see the solution.

Solution.

We apply the Fundamental Theorem of Calculus, Part (1:)

[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]

Hence

[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_1^x {sqrt {{t^3} + 4t} dt} } right) }={ sqrt {{x^3} + 4x} .}]

Substituting (x = 2) yields

[{g^primeleft( 2 right) }={ sqrt {{2^3} + 4 cdot 2} }={ sqrt {16} }={ 4.}]

Solution.

We use the Fundamental Theorem of Calculus, Part (1:)

[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]

Then

[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_{ – frac{pi }{2}}^x {sqrt {{{sin }^2}t + 2} dt} } right) }={ sqrt {{{sin }^2}x + 2} .}]

Note that the lower limit of integration ({ – large{frac{pi }{2}}normalsize}) does not affect the answer.

Now we compute the value of the derivative for (x = large{frac{pi }{6}}normalsize :)

[{g^primeleft( {frac{pi }{6}} right) }={ sqrt {{{sin }^2}frac{pi }{6} + 2} }={ sqrt {{{left( {frac{1}{2}} right)}^2} + 2} }={ sqrt {frac{9}{4}} }={ frac{3}{2}.}]

Solution.

We introduce the new function

[{hleft( u right) = intlimits_3^u {frac{{dt}}{t}}.}]

Using the FTC1, we have

[{h^primeleft( u right) }={ frac{1}{u}.}]

As (gleft( x right) = hleft( {{x^2}} right),) then by the chain rule

[{g^primeleft( x right) = left[ {hleft( {{x^2}} right)} right]^prime }={ h^primeleft( {{x^2}} right) cdot left( {{x^2}} right)^prime }={ h^primeleft( {{x^2}} right) cdot 2x }={ frac{1}{{{x^2}}} cdot 2x }={ frac{2}{x}}]

8.2first Fundamental Theorem Of Calculus Ap Calculus 2

Solution.

Since the upper limit of integration is not (x,) we apply the chain rule. Let (u = {x^2},) then (u^prime = 2x.)

Consider the new function

[hleft( u right) = intlimits_0^u {sqrt {1 + {t^2}} dt} .]

By the FTC1, we can write

[h^primeleft( u right) = sqrt {1 + {u^2}} .]

As (gleft( x right) = hleft( {{x^2}} right),) we have

[{g^primeleft( x right) = left[ {hleft( {{x^2}} right)} right]^prime }={ h^primeleft( {{x^2}} right) cdot left( {{x^2}} right)^prime }={ sqrt {1 + {{left( {{x^2}} right)}^2}} cdot 2x }={ 2xsqrt {1 + {x^4}} .}]

Solution.

Let (u = {x^3},) then (u^prime = 3{x^2}.)

We introduce the new function

[hleft( u right) = intlimits_0^u {{t^2}dt} .]

Using the FTC1, we obtain

[h^primeleft( u right) = {u^2}.]

Since (gleft( x right) = hleft( {{x^3}} right),) we have

[{g^primeleft( x right) = left[ {hleft( {{x^3}} right)} right]^prime }={ h^primeleft( {{x^3}} right) cdot left( {{x^3}} right)^prime }={ {left( {{x^3}} right)^2} cdot 3{x^2} }={ {x^6} cdot 3{x^2} }={ 3{x^8}.}]

Solution.

We split the interval of integration (left[ {{x^2},{x^3}} right]) using an intermediate point (c,) so that (c in left[ {{x^2},{x^3}} right].) Hence the derivative of (gleft( x right)) is written in the form

[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_{{x^2}}^{{x^3}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_{{x^2}}^c {tdt} + intlimits_c^{{x^3}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} – intlimits_c^{{x^2}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} } right) – frac{d}{{dx}}left( {intlimits_c^{{x^2}} {tdt} } right).}]

We calculate both terms using the FTC1 and the chain rule:

[{frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} } right) }={ {x^3} cdot left( {{x^3}} right)^prime }={ {x^3} cdot 3{x^2} }={ 3{x^5};}]

[{frac{d}{{dx}}left( {intlimits_c^{{x^2}} {tdt} } right) }={ {x^2} cdot left( {{x^2}} right)^prime }={ {x^2} cdot 2x }={ 2{x^3}.}]

Then

[g^primeleft( x right) = 3{x^5} – 2{x^3}.]

8.2first Fundamental Theorem Of Calculus Ap Calculus Answers

Solution.

We split the integral function into two terms:

[{gleft( x right) }={ intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} }={ intlimits_{sqrt x }^c {left( {{t^2} – t} right)dt} + intlimits_c^x {left( {{t^2} – t} right)dt} }={ intlimits_c^x {left( {{t^2} – t} right)dt} – intlimits_c^{sqrt x } {left( {{t^2} – t} right)dt},}]

where (c in left[ {{x^2},{x^3}} right].)

Find the derivative of (gleft( x right)) using the FTC1 and the chain rule (for the second term):

[{frac{d}{{dx}}intlimits_c^x {left( {{t^2} – t} right)dt} }={ {x^2} – x;}]

[{frac{d}{{dx}}intlimits_c^{sqrt x } {left( {{t^2} – t} right)dt} }={ left( {{{left( {sqrt x } right)}^2} – sqrt x } right) cdot left( {sqrt x } right)^prime }={ left( {x – sqrt x } right) cdot frac{1}{{2sqrt x }} }={ frac{{sqrt x }}{2} – frac{1}{2}.}]

Then

[{g^primeleft( x right) }={ left( {{x^2} – x} right) }-{ left( {frac{{sqrt x }}{2} – frac{1}{2}} right) }={ {x^2} – x – frac{{sqrt x }}{2} + frac{1}{2}.}]

At the point (x = 1,) the derivative is equal to

[{g^primeleft( 1 right) }={ {1^2} – 1 }-{ frac{{sqrt 1 }}{2} + frac{1}{2} }={ 0.}]

Solution.

Using the Fundamental Theorem of Calculus, Part (2,) we have

[
{intlimits_0^2 {left( {{x^3} – {x^2}} right)dx} }
= {left. {left( {frac{{{x^4}}}{4} – frac{{{x^3}}}{3}} right)} right|_0^2 }
= {left( {frac{{16}}{4} – frac{8}{3}} right) – 0 }={ frac{4}{3}.}
]

Solution.

An antiderivative of the function ({{t^2} + {t^{21}}}) is (large{frac{{{t^3}}}{3} + frac{{{t^{22}}}}{{22}}}normalsize.) Then using the Fundamental Theorem of Calculus, Part (2,) we have

[require{cancel}{intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt} }={ left. {left[ {frac{{{t^3}}}{3} + frac{{{t^{22}}}}{{22}}} right]} right|_{ – 1}^1 }={ left( {frac{{{1^3}}}{3} + frac{{{1^{22}}}}{{22}}} right) }-{ left( {frac{{{{left( { – 1} right)}^3}}}{3} + frac{{{{left( { – 1} right)}^{22}}}}{{22}}} right) }={ frac{1}{3} + cancel{frac{1}{{22}}} + frac{1}{3} – cancel{frac{1}{{22}}} }={ frac{2}{3}.}]

Solution.

[
{intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt} }
= {intlimits_0^1 {left( {{t^{largefrac{1}{3}normalsize}} – {t^{largefrac{1}{2}normalsize}}} right)dt} }
= {left. {left( {frac{{{t^{largefrac{1}{3}normalsize + 1}}}}{{frac{1}{3} + 1}} – frac{{{t^{largefrac{1}{2}normalsize + 1}}}}{{frac{1}{2} + 1}}} right)} right|_0^1 }
= {left. {left( {frac{{3{t^{largefrac{4}{3}normalsize}}}}{4} – frac{{2{t^{largefrac{3}{2}normalsize}}}}{3}} right)} right|_0^1 }
= {left( {frac{3}{4} – frac{2}{3}} right) – 0 }={ frac{1}{{12}}.}
]

Solution.

First we make the substitution:

[
{t = 3{x^2} – 1,;;}Rightarrow
{dt = 6xdx,;;}Rightarrow
{xdx = frac{{dt}}{6}.}
]

Determine the new limits of integration. When (x = 0,) then (t = -1.) When (x = 1,) then we have (t = 2.) So, the integral with the new variable (t) can be easily calculated:

[
{intlimits_0^1 {frac{x}{{{{left( {3{x^2} – 1} right)}^4}}}dx} }
= {intlimits_{ – 1}^2 {frac{{frac{{dt}}{6}}}{{{t^4}}}} }
= {frac{1}{6}int {{t^{ – 4}}dt} }
= {frac{1}{6}left. {left( {frac{{{t^{ – 3}}}}{{ – 3}}} right)} right|_{ – 1}^2 }
= { – frac{1}{{18}}left( {frac{1}{8} – 1} right) }
= {frac{7}{{144}}.}
]

Solution.

An antiderivative of the function ({t + large{frac{1}{t}}normalsize}) has the form (large{frac{{{t^2}}}{2}}normalsize + ln t.) Hence, by the Fundamental Theorem, Part (2,) we have

[{intlimits_1^e {left( {t + frac{1}{t}} right)dt} }={ left. {left[ {frac{{{t^2}}}{2} + ln t} right]} right|_1^e }={ left( {frac{{{e^2}}}{2} + ln e} right) }-{ left( {frac{{{1^2}}}{2} + ln 1} right) }={ frac{{{e^2}}}{2} + 1 – frac{1}{2} – 0 }={ frac{{{e^2}}}{2} + frac{1}{2}.}]

Solution.

We can write

[
{I = intlimits_0^{ln 2} {x{e^{ – x}}dx} }
= { – intlimits_0^{ln 2} {xdleft( {{e^{ – x}}} right)} .}
]

Apply integration by parts: ({largeintnormalsize} {udv} ) (= uv – {largeintnormalsize} {vdu} .) In this case, let

[
{u = x,;;}kern-0.3pt{dv = dleft( {{e^{ – x}}} right),;;}Rightarrow
{du = 1,;;}kern-0.3pt{v = {e^{ – x}}.}
]

Hence, the integral is

[
{I = – intlimits_0^{ln 2} {xdleft( {{e^{ – x}}} right)} }
= { – left[ {left. {left( {x{e^{ – x}}} right)} right|_0^{ln 2} – intlimits_0^{ln 2} {{e^{ – x}}dx} } right] }
= {{ – left. {left( {x{e^{ – x}}} right)} right|_0^{ln 2} }+{ intlimits_0^{ln 2} {{e^{ – x}}dx} }}
= {{ – left. {left( {x{e^{ – x}}} right)} right|_0^{ln 2} }-{ left. {left( {{e^{ – x}}} right)} right|_0^{ln 2} }}
= { – left. {left[ {{e^{ – x}}left( {x + 1} right)} right]} right|_0^{ln 2} }
= {{ – {e^{ – ln 2}}left( {ln 2 + 1} right) }+{ {e^0} cdot 1 }}
= { – frac{{ln 2}}{2} – frac{{ln e}}{2} + ln e }
= {frac{{ln e}}{2} – frac{{ln 2}}{2} }
= {frac{1}{2}left( {ln e – ln 2} right) }
= {frac{1}{2}ln frac{e}{2}.}
]

Solution.

We rewrite the absolute value expression in the form

[left| {x – frac{1}{2}} right| = begin{cases}– x + frac{1}{2}, & text{if }x lt frac{1}{2}x – frac{1}{2}, & text{if }x ge frac{1}{2}end{cases}]

and split the interval of integration into two intervals such that

[{intlimits_{ – 1}^1 {left| {x – frac{1}{2}} right|dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left| {x – frac{1}{2}} right|dx} }+{ intlimits_{frac{1}{2}}^1 {left| {x – frac{1}{2}} right|dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left( { – x + frac{1}{2}} right)dx} }+{ intlimits_{frac{1}{2}}^1 {left( {x – frac{1}{2}} right)dx} .}]

Now we can apply the Fundamental Theorem of Calculus, Part (2,) to each of the integrals:

[{intlimits_{ – 1}^1 {left| {x – frac{1}{2}} right|dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left( { – x + frac{1}{2}} right)dx} }+{ intlimits_{frac{1}{2}}^1 {left( {x – frac{1}{2}} right)dx} }={ left[ { – frac{{{x^2}}}{2} + frac{x}{2}} right]_{ – 1}^{frac{1}{2}} }+{ left[ {frac{{{x^2}}}{2} – frac{x}{2}} right]_{frac{1}{2}}^1 }={ left[ {left( { – frac{1}{8} + frac{1}{4}} right) – left( { – frac{1}{2} – frac{1}{2}} right)} right] }+{ left[ {left( {cancel{frac{1}{2}} – cancel{frac{1}{2}}} right) – left( {frac{1}{8} – frac{1}{4}} right)} right] }={ frac{9}{8} + frac{1}{8} }={ frac{{10}}{8} }={ frac{5}{4}.}]

Solution.

We represent the absolute value expression as follows:

[{left| {{x^2} – 1} right| text{ = }}kern0pt{begin{cases}{{x^2} – 1}, & text{if }x in left( { – infty , – 1} right] cup left[ {1,infty } right)1 – {x^2}, & text{if }x in left( { – 1,1} right)end{cases}}]

So we can split the initial integral into two integrals:

[{intlimits_{ – 2}^1 {left| {{x^2} – 1} right|dx} }={ intlimits_{ – 2}^{ – 1} {left| {{x^2} – 1} right|dx} }+{ intlimits_{ – 1}^1 {left| {{x^2} – 1} right|dx} }={ intlimits_{ – 2}^{ – 1} {left( {{x^2} – 1} right)dx} }+{ intlimits_{ – 1}^1 {left( {1 – {x^2}} right)dx} .}]

Using the Fundamental Theorem of Calculus, Part (2,) we obtain:

[{intlimits_{ – 2}^1 {left| {{x^2} – 1} right|dx} }={ intlimits_{ – 2}^{ – 1} {left( {{x^2} – 1} right)dx} }+{ intlimits_{ – 1}^1 {left( {1 – {x^2}} right)dx} }={ left[ {frac{{{x^3}}}{3} – x} right]_{ – 2}^{ – 1} }+{ left[ {x – frac{{{x^3}}}{3}} right]_{ – 1}^1 }={ left[ {left( { – frac{1}{3} – left( { – 1} right)} right) }right.}-{left.{ left( { – frac{8}{3} – left( { – 2} right)} right)} right] }+{ left[ {left( {1 – frac{1}{3}} right) – left( { – 1 – left( { – frac{1}{3}} right)} right)} right] }={ frac{2}{3} + frac{2}{3} + frac{2}{3} + frac{2}{3} }={ frac{8}{3}.}]

Solution.

First we find the points of intersection (see Figure (6)):

[
{{x^2} = sqrt x ,;;}Rightarrow
{{x^2} – sqrt x = 0,;;}Rightarrow
{sqrt x left( {{x^{largefrac{3}{2}normalsize}} – 1} right) = 0,;;}Rightarrow
{{x_1} = 0,;{x_2} = 1.}
]

As you can see, the curves intercept at the points (left( {0,0} right)) and (left( {1,1}right).) Hence, the area is given by

[
{S = intlimits_0^1 {left( {sqrt x – {x^2}} right)dx} }
= {left. {left( {frac{{{x^{largefrac{1}{2}normalsize + 1}}}}{{frac{1}{2} + 1}} – frac{{{x^3}}}{3}} right)} right|_0^1 }
= {frac{1}{3}left. {left( {2sqrt {{x^3}} – {x^3}} right)} right|_0^1 }={ frac{1}{3}.}
]

Solution.

First we find the points of intersection of the curves (see Figure (7)):

[
{2x – {x^2} = – x,;;}Rightarrow
{{x^2} – 3x = 0,;;}Rightarrow
{xleft( {x – 3} right) = 0,;;}Rightarrow
{{x_1} = 0,;{x_2} = 3.}
]

The upper boundary of the region is the parabola (y = 2x – {x^2},) and the lower boundary is the straight line (y = -x.)

The area is given by

[
{S = intlimits_0^3 {left[ {2x – {x^2} – left( { – x} right)} right]dx} }
= {intlimits_0^3 {left( {2x – {x^2} + x} right)dx} }
= {left. {left( {{x^2} – frac{{{x^3}}}{3} + frac{{{x^2}}}{2}} right)} right|_0^3 }
= {left. {left( {frac{{3{x^2}}}{2} – frac{{{x^3}}}{3}} right)} right|_0^3 }
= {frac{{27}}{2} – frac{{27}}{3} }={ frac{9}{2}.}
]

Solution.

First we find an equation of the side (OA) (Figure (8)):

[
{frac{{x – {x_O}}}{{{x_A} – {x_O}}} = frac{{y – {y_O}}}{{{y_A} – {y_O}}},;;}Rightarrow
{frac{{x – 0}}{{2 – 0}} = frac{{y – 0}}{{6 – 0}},;;}Rightarrow
{frac{x}{2} = frac{y}{6},;;}Rightarrow
{y = 3x.}
]

Similarly, we find an equation of the side (OB:)

[
{frac{{x – {x_O}}}{{{x_B} – {x_O}}} = frac{{y – {y_O}}}{{{y_B} – {y_O}}},;;}Rightarrow
{frac{{x – 0}}{{7 – 0}} = frac{{y – 0}}{{1 – 0}},;;}Rightarrow
{frac{x}{7} = frac{y}{1},;;}Rightarrow
{y = frac{x}{7}.}
]

Next, find an equation of the side (AB:)

[
{frac{{x – {x_B}}}{{{x_A} – {x_B}}} = frac{{y – {y_B}}}{{{y_A} – {y_B}}},;;}Rightarrow
{frac{{x – 2}}{{7 – 2}} = frac{{y – 6}}{{1 – 6}},;;}Rightarrow
{frac{{x – 2}}{5} = frac{{y – 6}}{{ – 5}},;;}Rightarrow
{y = 8 – x.}
]

As you can see from Figure (8,) the area of the this triangle can be calculated as the sum of two integrals:

[
{S = {I_1} + {I_2} }
= {intlimits_0^2 {left( {3x – frac{x}{7}} right)dx} }+{ intlimits_2^7 {left( {8 – x – frac{x}{7}} right)dx} }
= {left. {left( {frac{{10{x^2}}}{7}} right)} right|_0^2 }+{ left. {left( {8x – frac{{4{x^2}}}{7}} right)} right|_2^7 }
= {frac{{10 cdot 4}}{7} + left( {56 – frac{{4 cdot 49}}{7}} right) }-{ left( {16 – frac{{4 cdot 4}}{7}} right) }={ 20.}
]

Solution.

By symmetry (see Figure (9)), the area of the ellipse is twice the area above the (x)-axis.

The latter is given by

[
{{S_{frac{1}{2}}} }={ intlimits_{ – a}^a {sqrt {{b^2}left( {1 – frac{{{x^2}}}{{{a^2}}}} right)} dx} }
= {frac{b}{a}intlimits_{ – a}^a {sqrt {{a^2} – {x^2}} dx} .}
]

To calculate the last integral, we use the trigonometric substitution (x = asin t,) (dx = acos tdt.)

Refine the limits of integration. When (x = -a,) then (sin t = -1) and (t = – {largefrac{pi }{2}normalsize}.) When (x = a,) then (sin t = 1) and (t = {largefrac{pi }{2}normalsize}.) Thus we get

[
{{S_{frac{1}{2}}} }={ frac{b}{a}intlimits_{ – a}^a {sqrt {{a^2} – {x^2}} dx} }
= {frac{b}{a}intlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {sqrt {{a^2} – {a^2}{{sin }^2}t}, }}kern0pt{{ acos tdt} }
= {abintlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {{{cos }^2}tdt} }
= {abintlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {frac{{1 + cos 2t}}{2}dt} }
= {frac{{ab}}{2}intlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {left( {1 + cos 2t} right)dt} }
= {frac{{ab}}{2}left. {left( {t + frac{{sin 2t}}{2}} right)} right|_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} }
= {frac{{ab}}{2}left[ {frac{pi }{2} + frac{{sin pi }}{2} }right.}-{left.{ left( { – frac{pi }{2}} right) – frac{{sin left( { – pi } right)}}{2}} right] }
= {frac{{pi ab}}{2}.}
]

Hence, the total area of the ellipse is (pi ab.)

Many people believe that mathematics is about number-crunching, but much more importantly, math is about reasoning. For example, when you solve a word problem, you are using your reasoning skills to put together the given information in just the right way.

In a way, AP Calculus is all about reasoning. You have to interpret each problem and correctly apply the appropriate methods (limits, derivatives, integrals, etc.) to solve it. However sometimes we have to take it one step further and reason with theorems and definitions as well, gluing our thoughts together with mathematical logic.

Why is this important? Well using nothing more than a handful of assumptions and plenty of definitions, theorems, and logic, Euclid developed the entire subject of Geometry from the ground up! If that’s not a reason to respect the power of definitions and theorems, then nothing else is.

What are Definitions and Theorems?

In mathematics, every term must be defined in some way. A definition of a mathematical object is formal description of the essential properties that make that object what it is. For instance,

• Definition: A triangle is a three-sided polygon.

It’s very important to understand the definitions of our mathematical terms so that we can employ just the right tool in each specific case. So if you see a three-sided polygon in a problem, then you know that it’s a triangle by definition. Then you may use a property or formula related to triangles as part of your reasoning steps.

We also rely on general statements of truth called theorems in order to reason about a specific situation. Speaking of triangles, perhaps one of the most famous (and useful) theorems of all time is the Pythagorean Theorem. (By the way, this theorem shows up in Book 1 of Euclid’s Elements, over 2000 years ago!)

Diagram for Pythagoras theorem by Drini (Pedro Sanchez)

Reasoning with Definitions

On the AP Calculus exams, you must know and be able to apply the definitions of calculus. Let’s see what that means in an example problem.

Example 1

If , which of the following is true?

(A) f(x) is continuous and differentiable at x = 3

(B) f(x) is continuous but not differentiable at x = 3

(C) f(x) is neither continuous nor differentiable at x = 3

(D) f(x) is differentiable but not continuous at x = 3

In order to properly address this question, we must know the definitions of continuous and differentiable.

• A function f is continuous at a point x = a if
• A function f is differentiable at a point x = a if the derivative f ‘(a) exists.

Example Solution — Continuity

First let’s determine if the function is continuous at x = 3. Because f is defined piece-wise, we must compute both the left and right hand limits.

Now because the left and right hand limits agree, we know that the two-sided limit as x → 3 exists and equals 0. Next, check the function value at x = 3.

Therefore, since the limiting value equals the function value (both are 0), the function f is continuous at x = 3 by definition. (For more about this topic, check out AP Calculus Exam Review: Limits and Continuity.)

Example Solution — Differentiability

Moving on to differentiability, now we must check whether f ‘(3) exists. Again, because f is defined piece-wise, we must be careful at the point where the function changes behavior. First find the derivative of each piece.

Note, there is no typo here — the derivative of the first piece can only be found when x < 3. In fact it takes more analysis to figure out what happens atx = 3.

Because the derivative itself is actually a certain kind of limit (by definition!), we’ll have to see what the limiting values for f ‘ are as x → 3. As before, examine each piece separately.

This time there is a mismatch. Because the left and right derivatives do not agree (18 ≠ -9), the derivative does not exist at x = 3. Thus by definition, f is not differentiable at x = 3.

In summary, f is continuous, but not differentiable at x = 3. Choice (B) is correct.

Reasoning with Theorems

Remember, a theorem is a true mathematical statement. Typically theorems are general facts that can apply to lots of different situations. Here is a small list of important theorems in calculus.

• Intermediate Value Theorem
• Extreme Value Theorem
• Mean Value Theorem for Derivatives
• Rolle’s Theorem
• Fundamental Theorem of Calculus (two parts)
• Mean Value Theorem for Integrals

A Theorem by any other Name…

There are many other results and formulas in calculus that may not have the title of “Theorem” but are nevertheless important theorems. Every one of your derivative and antidifferentiation rules is actually a theorem. Here is a partial list of other theorems that may not be explicitly identified as theorems in your textbook.

• Differentiability implies continuity
• The first derivative rule for increase and decrease
• The second derivative rule for concavity
• First and second derivative rules for relative extrema
• Product Rule, Quotient Rule, Chain Rule, etc.
• Additivity and linearity of the definite integral
• Techniques of antidifferentiation such as substitution, integration by parts, etc.
• Various tests for convergence of series

Now let’s see if we can use the right theorems to crack the next example.

Example 2

Find .

Notice that this is a derivative of an integral. That means we may be able to apply the Fundamental Theorem of Calculus. There are two parts to the theorem, but the one we need is:

However, before we can apply this theorem, we must change the form of the integral. The theorem requires that the lower limit of integration must be a constant. By using the rule for switching the order of integration (another theorem!), we may write:

Next, because the upper limit of integration is not a simple variable, x, we must use yet another theorem: the Chain Rule. Here, the “inside function” is u = x³.

It’s interesting to note in this case that no other method could have led to the solution. It is impossible to write down an antiderivative for the function, sin t². Fortunately the Fundamental Theorem of Calculus in the form we used it avoids the antidifferentiation step altogether.

Conclusion

Definitions and theorems form the backbone of mathematical reasoning. Knowing your definitions means knowing which tools can apply in each situation. And by understanding the theorems, you can avoid doing a lot of unnecessary or difficult work.

8.2first Fundamental Theorem Of Calculus Ap Calculus Pdf

About Shaun Ault

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

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