When Can We Change Signs Of The Limits Of An Integral

4.7: Definite integrals by exchange.

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Substitution for Definite Integrals

Exchange can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration alter as well.

Substitution with Definite Integrals

Permit $u=g(10)$ and let $g'$ exist continuous over an interval $[a,b]$, and permit $f$ be continuous over the range of $u=g(ten).$ Then,

\[∫^b_af(chiliad(x))g′(x)dx=∫^{g(b)}_{g(a)}f(u)\,du.\]

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $F(x)$ is an antiderivative of $f(10),$ we take

\[ ∫f(1000(10))g′(ten)\,dx=F(thou(x))+C.\]

Then

\[\brainstorm{align} ∫^b_af[grand(x)]m′(x)\,dx &= F(m(10))\bigg|^{x=b}_{10=a} \nonumber \\ &=F(m(b))−F(chiliad(a)) \nonumber\\ &= F(u) \bigg|^{u=g(b)}_{u=thou(a)} \nonumber\\ &=∫^{yard(b)}_{g(a)}f(u)\,du \nonumber\finish{marshal} \nonumber\]

and nosotros have the desired result.

Instance $\PageIndex{5}$: Using Exchange to Evaluate a Definite Integral

Use substitution to evaluate \[ ∫^1_0x^two(one+2x^three)^five\,dx.\]

Solution

Let $u=ane+2x^3$, so $du=6x^2dx$. Since the original part includes one gene of $ten^2$ and $du=6x^2dx$, multiply both sides of the du equation by $one/vi.$ And then,

\[ du=6x^ii\,dx\]

\[ \dfrac{1}{6}du=10^2\,dx.\]

To adjust the limits of integration, annotation that when $x=0,u=one+2(0)=1,$ and when $x=1,u=1+2(1)=three.$ Then

\[ ∫^1_0x^ii(1+2x^3)^5dx=\dfrac{i}{vi}∫^3_1u^5\,du.\]

Evaluating this expression, we get

\[ \dfrac{1}{6}∫^3_1u^5\,du=(\dfrac{1}{vi})(\dfrac{u^6}{6})|^3_1=\dfrac{1}{36}[(3)^vi−(1)^6]=\dfrac{182}{9}.\]

Exercise $\PageIndex{5}$

Use substitution to evaluate the definite integral \[ ∫^0_{−i}y(2y^2−3)^5\,dy.\]

Hint: Use the steps from Example to solve the trouble.

Answer: $\dfrac{91}{3}$

Instance $\PageIndex{vi}$: Using Exchange with an Exponential Function

Use substitution to evaluate \[ ∫^1_0xe^{4x^2+3}\,dx.\]

Solution

Let $u=4x^three+3.$ Then, $du=8x\,dx.$ To adjust the limits of integration, nosotros annotation that when $ten=0,u=iii$, and when $ten=1,u=7$. So our commutation gives

\[ ∫^1_0xe^{4x^2+3}\,dx=\dfrac{1}{viii}∫^7_3e^udu=\dfrac{one}{eight}e^u|^7_3=\dfrac{e^7−e^3}{8}≈134.568\]

Exercise $\PageIndex{6}$

Use exchange to evaluate \[ ∫^1_0x^2cos(\dfrac{π}{ii}x^3)\,dx.\]

Hint: Use the process from Case to solve the problem.

Answer: $\dfrac{ii}{3π}≈0.2122$

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can utilize substitution. Besides, nosotros have the option of replacing the original expression for u after we discover the antiderivative, which means that we exercise not have to modify the limits of integration. These two approaches are shown in Instance.

Instance $\PageIndex{7}$: Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate \[∫^{π/2}_0\cos^2θ\,dθ.\]

Solution

Let u.s.a. start use a trigonometric identity to rewrite the integral. The trig identity $\cos^2θ=\dfrac{1+\cos 2θ}{2}$ allows us to rewrite the integral as

\[∫^{π/two}_0\cos^2θdθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ.\]

Then,

\[∫^{π/2}_0(\dfrac{ane+\cos2θ}{2})dθ=∫^{π/2}_0(\dfrac{1}{2}+\dfrac{1}{two}\cos 2θ)\,dθ\]

\[=\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ.\]

We can evaluate the start integral equally it is, but nosotros need to make a substitution to evaluate the 2nd integral. Let $u=2θ.$ And so, $du=2\,dθ,$ or $\dfrac{one}{2}\,du=dθ$. Also, when $θ=0,u=0,$ and when $θ=π/2,u=π.$ Expressing the second integral in terms of $u$, we have

$\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0cos^2θ\,dθ=\dfrac{1}{2}∫^{π/ii}_0\,dθ+\dfrac{1}{2}(\dfrac{1}{2})∫^π_0\cos u \,du$

$=\dfrac{θ}{ii}|^{θ=π/2}_{θ=0}+\dfrac{1}{4}sinu|^{u=θ}_{u=0}$

$=(\dfrac{π}{iv}−0)+(0−0)=\dfrac{π}{iv}$

Example $ \PageIndex{8}$: Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral

\[ ∫^1_0\dfrac{dx}{\sqrt{1−x^2}}. \nonumber\]

Solution

Nosotros tin become directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have

\[ ∫^1_0\dfrac{dx}{\sqrt{1−x^2}}=\sin^{−1}x∣^1_0=\sin^{−1}1−\sin^{−1}0=\dfrac{π}{ii}−0=\dfrac{π}{2}.\nonumber\

Case $ \PageIndex{9}$: Evaluating a Definite Integral

Evaluate the definite integral $ ∫^{\sqrt{three}}_{\sqrt{3}/3}\dfrac{dx}{i+x^2}$.

Solution

Employ the formula for the inverse tangent. We have

\[ ∫^{\sqrt{three}}_{\sqrt{3}/iii}\dfrac{dx}{1+ten^2}=tan^{−ane}x∣^{\sqrt{3}}_{\sqrt{three}/3} =[tan^{−1}(\sqrt{3})]−[tan^{−1}(\dfrac{\sqrt{3}}{3})]=\dfrac{π}{half-dozen}.\]

Exercise $\PageIndex{9}$

Evaluate the definite integral $ ∫^2_0\dfrac{dx}{4+x^2}$.

Hint: Follow the procedures from Example to solve the problem.

Answer: \[ \dfrac{π}{8}\]

Equally mentioned at the first of this section, exponential functions are used in many real-life applications. The number e is often associated with compounded or accelerating growth, equally we have seen in before sections about the derivative. Although the derivative represents a charge per unit of change or a growth rate, the integral represents the total change or the total growth. Allow's wait at an example in which integration of an exponential function solves a common business application.

A cost–demand function tells us the human relationship between the quantity of a production demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal toll–demand role is the derivative of the price–demand function and it tells u.s. how fast the price changes at a given level of production. These functions are used in business to determine the cost–elasticity of need, and to help companies determine whether irresolute production levels would be profitable.

Example $\PageIndex{4}$: Finding a Price–Demand Equation

Find the cost–need equation for a particular brand of toothpaste at a supermarket chain when the demand is fifty tubes per week at $2.35 per tube, given that the marginal cost—demand role, $p′(10),$ for x number of tubes per week, is given as

\[p'(10)=−0.015e^{−0.01x}.\]

If the supermarket chain sells 100 tubes per week, what price should it set?

Solution

To discover the price–demand equation, integrate the marginal price–demand function. Get-go detect the antiderivative, then look at the particulars. Thus,

\[p(ten)=∫−0.015e^{−0.01x}dx=−0.015∫east^{−0.01x}dx.\]

Using substitution, let $u=−0.01x$ and $du=−0.01dx$. And then, separate both sides of the du equation by −0.01. This gives

\[\dfrac{−0.015}{−0.01}∫e^udu=1.5∫e^udu=1.5e^u+C=1.5e^{−0.01}x+C.\]

The next stride is to solve for C. We know that when the toll is $ii.35 per tube, the need is 50 tubes per week. This means

\[p(50)=1.5e^{−0.01(50)}+C=ii.35.\]

At present, just solve for C:

\[C=2.35−i.5e^{−0.5}=2.35−0.91=1.44.\]

Thus,

\[p(x)=1.5e^{−0.01x}+1.44.\]

If the supermarket sells 100 tubes of toothpaste per calendar week, the price would exist

\[p(100)=i.5e−0.01(100)+ane.44=ane.5e−1+i.44≈1.99.\]

The supermarket should accuse $one.99 per tube if it is selling 100 tubes per week.

Case $\PageIndex{5}$: Evaluating a Definite Integral Involving an Exponential Function

Evaluate the definite integral \[∫^2_1e^{1−x}dx.\]

Solution

Once again, exchange is the method to utilize. Let $u=i−x,$ so $du=−1dx$ or $−du=dx$. Then $∫e^{1−x}dx=−∫due east^udu.$ Next, modify the limits of integration. Using the equation $u=1−ten$, we accept

\[u=i−(1)=0\]

\[u=1−(ii)=−one.\]

The integral so becomes

\[∫^2_1e^{i−x}\,dx=−∫^{−i}_0e^u\,du=∫^0_{−one}e^u\,du=eu|^0_{−1}=eastward^0−(eastward^{−1})=−e^{−1}+one.\]

Encounter Figure.

alt — Figure $\PageIndex{2}$: The indicated area tin be calculated past evaluating a definite integral using substitution.

Exercise $\PageIndex{four}$

Evaluate $∫^2_0e^{2x}dx.$

Hint: Let $u=2x.$

Answer: $\dfrac{1}{2}∫^4_0e^udu=\dfrac{1}{two}(e^4−1)$

Case $\PageIndex{6}$: Growth of Leaner in a Civilisation

Suppose the charge per unit of growth of bacteria in a Petri dish is given by $q(t)=3^t$, where t is given in hours and $q(t)$ is given in thousands of bacteria per 60 minutes. If a civilisation starts with 10,000 leaner, detect a function $Q(t)$ that gives the number of bacteria in the Petri dish at whatever time t. How many bacteria are in the dish after 2 hours?

Solution

We have

\[Q(t)=∫iii^tdt=\dfrac{three^t}{\ln iii}+C.\]

So, at $t=0$ we have $Q(0)=10=\dfrac{1}{\ln 3}+C,$ so $C≈9.090$ and we go

\[Q(t)=\dfrac{3^t}{\ln 3}+9.090.\]

At time $t=2$, we have

\[Q(2)=\dfrac{3^2}{\ln 3}+nine.090\]

\[=17.282.\]

After two hours, there are 17,282 bacteria in the dish.

Practice $\PageIndex{5}$

From Instance, suppose the bacteria grow at a rate of $q(t)=2^t$. Assume the culture still starts with 10,000 bacteria. Discover $Q(t)$. How many leaner are in the dish after three hours?

Hint: Use the process from Example to solve the problem

Reply: $Q(t)=\dfrac{2^t}{\ln 2}+8.557.$ In that location are 20,099 bacteria in the dish after 3 hours.

Example $\PageIndex{seven}$: Fruit Fly Population Growth

Suppose a population of fruit flies increases at a rate of $g(t)=2e^{0.02t}$, in flies per mean solar day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days?

Solution

Allow $G(t)$ represent the number of flies in the population at time t. Applying the net modify theorem, nosotros accept

$Thou(10)=Grand(0)+∫^{10}_02e^{0.02t}dt$

$=100+[\dfrac{2}{0.02}east^{0.02t}]∣^{10}_0$

$=100+[100e^{0.02t}]∣^{ten}_0$

$=100+100e^{0.2}−100$

$≈122.$

At that place are 122 flies in the population after ten days.

Exercise $\PageIndex{vi}$

Suppose the rate of growth of the wing population is given by $one thousand(t)=e^{0.01t},$ and the initial fly population is 100 flies. How many flies are in the population after fifteen days?

Hint: Use the process from Case to solve the problem.

Answer: There are 116 flies.

Example $\PageIndex{8}$: Evaluating a Definite Integral Using Substitution

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{east^{ane/10}}{x^2}\,dx.\]

Solution

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x, then bring the $x^2$ in the denominator upward to the numerator using a negative exponent. We have

\[∫^2_1\dfrac{e^{1/x}}{x^2}\,dx=∫^2_1e^{x^{−1}}10^{−ii}\,dx.\]

Allow $u=x^{−1},$ the exponent on $eastward$. Then

\[du=−x^{−2}\,dx\]

\[−du=ten^{−2}\,dx.\]

Bringing the negative sign exterior the integral sign, the trouble now reads

\[−∫e^u\,du.\]

Next, alter the limits of integration:

\[u=(ane)^{−ane}=i\]

\[u=(ii)^{−1}=\dfrac{1}{two}.\]

Observe that now the limits begin with the larger number, meaning nosotros must multiply by −1 and interchange the limits. Thus,

\[−∫^{1/two}_1e^udu=∫^1_{one/two}eastward^udu=e^u|^1_{1/2}=e−e^{ane/2}=e−\sqrt{eastward}.\]

Do $\PageIndex{seven}$

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{1}{x^three}due east^{4x^{−ii}}dx.\]

Hint: Permit $u=4x^{−2}.$

Answer: \[∫^2_1\dfrac{1}{x^3}eastward^{4x^{−2}}dx=\dfrac{1}{8}[e^4−e]\].

Example is a definite integral of a trigonometric role. With trigonometric functions, we frequently have to employ a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is unremarkably the fundamental to a smooth integration.

Example $\PageIndex{12}$: Evaluating a Definite Integral

Find the definite integral of \[∫^{π/2}_0\dfrac{\sin x}{1+\cos x}dx.\]

Solution

We need commutation to evaluate this problem. Let $u=1+\cos x$ so $du=−\sin x\,dx.$ Rewrite the integral in terms of u, changing the limits of integration besides. Thus,

\[u=1+cos(0)=two\]

\[u=i+cos(\dfrac{π}{ii})=ane.\]

Then

\[∫^{π/2}_0\dfrac{\sin x}{1+\cos x}=−∫^one+2u^{−ane}du=∫^2_1u^{−1}du=\ln |u|^2_1=[\ln ii−\ln 1]=\ln ii\]

Fundamental Concepts

Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term 'substitution' refers to irresolute variables or substituting the variable u and du for appropriate expressions in the integrand.
When using substitution for a definite integral, we too have to change the limits of integration.

Primal Equations

Substitution with Definite Integrals

$∫^b_af(m(x))g'(x)dx=∫^{yard(b)}_{grand(a)}f(u)du$

Glossary

change of variables: the substitution of a variable, such as u, for an expression in the integrand

integration by exchange: a technique for integration that allows integration of functions that are the issue of a chain-rule derivative

Contributors and Attributions

Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for gratis at http://cnx.org.

Source: https://math.libretexts.org/Courses/Mount_Royal_University/MATH_1200%3A_Calculus_for_Scientists_I/4%3A_Integral_Calculus/4.7%3A_Definite_integrals_by_substitution.

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