# 8.2first Fundamental Theorem Of Calculusap Calculus

Following are some of the most frequently used theorems, formulas, and definitions that you encounter in a calculus class for a single variable. The list isn’t comprehensive, but it should cover the items you’ll use most often.

## Definition of a Critical Number

A critical number of a function f is a number c in the domain of f such that either f(c)= 0 or f(c) does not exist.

Following are some of the most frequently used theorems, formulas, and definitions that you encounter in a calculus class for a single variable. The list isn’t comprehensive, but it should cover the items you’ll use most often. Limit Definition of a Derivative Definition: Continuous at a number a The Intermediate Value Theorem Definition of a. The first part of the fundamental theorem stets that when solving indefinite integrals between two points a and b, just subtract the value of the integral at a from the value of the integral at b. The second part of the theorem gives an indefinite integral of a function. Fundamental Theorem of Calculus Solutions. We have intentionally included more material than can be covered in most Student Study Sessions to account for groups that are able to answer the questions at a faster rate. Use your own judgment, based on the group of students, to determine the order and selection of questions to work in the session. PETERSON’S MASTER AP CALCULUS AB&BC 2nd Edition W. Michael Kelley Mark Wilding, Contributing Author.

3 fails to meet either of these requirements. To apply the Fundamental Theorem of Calculus, the symbolic integration utility must be able to find the antiderivative. Which method would you use to evaluate Which method would you use to evaluate 1 0 arctan x2 dx? 1 0 arctan x dx? 3.qxd 11/2/04 3:05 PM Page 527.

## Rolle’s Theorem

Let f be a function that satisfies the following three hypotheses:

• f is continuous on the closed interval [a, b].

• f is differentiable on the open interval (a, b).

• f(a)= f(b).

Then there is a number c in (a, b) such that f(c)= 0.

## The Mean Value Theorem

Let f be a function that satisfies the following hypotheses:

• f is continuous on the closed interval [a, b].

• f is differentiable on the open interval (a, b).

## Newton’s Method Approximation Formula

Newton’s method is a technique that tries to find a root of an equation. To begin, you try to pick a number that’s “close” to the value of a root and call this value x1. Picking x1 may involve some trial and error; if you’re dealing with a continuous function on some interval (or possibly the entire real line), the intermediate value theorem may narrow down the interval under consideration. After picking x1, you use the recursive formula given here to find successive approximations:

A word of caution: Always verify that your final approximation is correct (or close to the value of the root). Newton’s method can fail in some instances, based on the value picked for x1. Any calculus text that covers Newton’s method should point out these shortcomings.

## The Fundamental Theorem of Calculus

Suppose f is continuous on [a, b]. Then the following statements are true:

## The Trapezoid Rule

where  ## Simpson’s Rule

where n is even and

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

### 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

## 8.2first Fundamental Theorem Of Calculus Ap Calculus Pdf

[{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.

### Example 1

Calculate the derivative of the function (gleft( x right) = intlimits_1^x {sqrt {{t^3} + 4t} dt} ) at (x = 2.)

### Example 2

Calculate the derivative of the function (gleft( x right) = intlimits_{ – large{frac{pi }{2}}normalsize}^x {sqrt {{{sin }^2}t + 2} dt} ) at (x = large{large{frac{pi }{6}}normalsize}.)

### Example 3

Find the derivative of the function (gleft( x right) = intlimits_3^{{x^2}} {large{frac{{dt}}{t}}normalsize}.) ### Example 4

Find the derivative of the function (gleft( x right) = intlimits_0^{{x^2}} {sqrt {1 + {t^2}} dt}.)

### Example 5

Find the derivative of the function (gleft( x right) = intlimits_1^{{x^3}} {{t^2}dt}.)

### Example 6

Find the derivative of the function (gleft( x right) = intlimits_{{x^2}}^{{x^3}} {tdt}.)

### Example 7

Calculate the derivative of the function (gleft( x right) = intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} ) at (x = 1.)

### Example 8

Evaluate the integral (intlimits_0^2 {left( {{x^3} – {x^2}} right)dx}.)

### Example 9

Evaluate the integral (intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt}.)

## 8.2first Fundamental Theorem Of Calculus Ap Calculus 2

### Example 10

Calculate the integral (intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt}.)

### Example 11

Evaluate the integral (intlimits_0^1 {{largefrac{x}{{{{left( {3{x^2} – 1} right)}^4}}}normalsize} dx}.)

### Example 12

Evaluate the integral (intlimits_1^e {left( {t + large{frac{1}{t}}normalsize} right)dt}.)

### Example 13

Evaluate the integral (intlimits_0^{ln 2} {x{e^{ – x}}dx}.)

### Example 14

Evaluate the integral (intlimits_{ – 1}^1 {left {x – large{frac{1}{2}}normalsize} right dx}.)

### Example 15

Evaluate the integral (intlimits_{ – 2}^1 {left {{x^2} – 1} right dx}.)

### Example 16

Find the area bounded by the curves (y = {x^2}) and (y = sqrt x.)

### Example 17

Find the area bounded by the curves (y = 2x – {x^2}) and (x + y = 0.)

### Example 18

Find the area of the triangle with vertices at (left( {0,0} right),) (left( {2,6} right)) and (left( {7,1} right).)

### Example 19

Find the area inside the ellipse ({largefrac{{{x^2}}}{{{a^2}}}normalsize} + {largefrac{{{y^2}}}{{{b^2}}}normalsize} = 1.)

### Example 1.

Calculate the derivative of the function (gleft( x right) = intlimits_1^x {sqrt {{t^3} + 4t} dt} ) at (x = 2.)

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

Convert dmg file to xd file. [{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.}]

### Example 2.

Calculate the derivative of the function (gleft( x right) = intlimits_{ – large{frac{pi }{2}}normalsize}^x {sqrt {{{sin }^2}t + 2} dt} ) at (x = large{large{frac{pi }{6}}normalsize}.)

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}.}]

### Example 3.

Find the derivative of the function (gleft( x right) = intlimits_3^{{x^2}} {large{frac{{dt}}{t}}normalsize}.)

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}}]

### Example 4.

Find the derivative of the function (gleft( x right) = intlimits_0^{{x^2}} {sqrt {1 + {t^2}} dt}.)

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}} .}]

### Example 5.

Find the derivative of the function (gleft( x right) = intlimits_1^{{x^3}} {{t^2}dt}.)

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}.}]

### Example 6.

Find the derivative of the function (gleft( x right) = intlimits_{{x^2}}^{{x^3}} {tdt}.)

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}.]

### Example 7.

Calculate the derivative of the function (gleft( x right) = intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} ) at (x = 1.)

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.}]

### Example 8.

Evaluate the integral (intlimits_0^2 {left( {{x^3} – {x^2}} right)dx}.)

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}.}
]

### Example 9.

Evaluate the integral (intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt}.)

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}.}]

### Example 10.

Calculate the integral (intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt}.)

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}}.}
]

### Example 11.

Evaluate the integral (intlimits_0^1 {{largefrac{x}{{{{left( {3{x^2} – 1} right)}^4}}}normalsize} dx}.)

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}}.}
]

### Example 12.

Evaluate the integral (intlimits_1^e {left( {t + large{frac{1}{t}}normalsize} right)dt}.)

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}.}]

### Example 13.

Evaluate the integral (intlimits_0^{ln 2} {x{e^{ – x}}dx}.)

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}.}
]

### Example 14.

Evaluate the integral (intlimits_{ – 1}^1 {left {x – large{frac{1}{2}}normalsize} right dx}.)

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}.}]

### Example 15.

Evaluate the integral (intlimits_{ – 2}^1 {left {{x^2} – 1} right dx}.)

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}.}]

## 8.2first Fundamental Theorem Of Calculus Ap Calculus Answers

### Example 16.

Find the area bounded by the curves (y = {x^2}) and (y = sqrt x.)

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}.}
]

### Example 17.

Find the area bounded by the curves (y = 2x – {x^2}) and (x + y = 0.)

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}.}
]

### Example 18.

Find the area of the triangle with vertices at (left( {0,0} right),) (left( {2,6} right)) and (left( {7,1} right).)

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.}
]

## 8.2first Fundamental Theorem Of Calculus Ap Calculus Free

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.}
]

### Example 19.

Find the area inside the ellipse ({largefrac{{{x^2}}}{{{a^2}}}normalsize} + {largefrac{{{y^2}}}{{{b^2}}}normalsize} = 1.)

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.)