When Can We Change Signs Of The Limits Of An Integral

four.7: Definite integrals by substitution.

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Exchange for Definite Integrals

Exchange can be used with definite integrals, besides. Withal, using substitution to evaluate a definite integral requires a modify to the limits of integration. If we modify variables in the integrand, the limits of integration change equally well.

Substitution with Definite Integrals

Let \(u=g(x)\) and permit \(m'\) be continuous over an interval \([a,b]\), and let \(f\) be continuous over the range of \(u=yard(x).\) And so,

\[∫^b_af(g(x))g′(10)dx=∫^{grand(b)}_{thousand(a)}f(u)\,du.\]

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

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

And so

\[\begin{align} ∫^b_af[g(10)]one thousand′(ten)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \nonumber \\ &=F(g(b))−F(thousand(a)) \nonumber\\ &= F(u) \bigg|^{u=g(b)}_{u=grand(a)} \nonumber\\ &=∫^{g(b)}_{thousand(a)}f(u)\,du \nonumber\end{align} \nonumber\]

and we have the desired result.

Instance \(\PageIndex{five}\): Using Substitution to Evaluate a Definite Integral

Utilise substitution to evaluate \[ ∫^1_0x^2(ane+2x^iii)^5\,dx.\]

Solution

Let \(u=one+2x^3\), and so \(du=6x^2dx\). Since the original role includes one factor of \(x^2\) and \(du=6x^2dx\), multiply both sides of the du equation by \(1/6.\) And then,

\[ du=6x^2\,dx\]

\[ \dfrac{one}{vi}du=x^2\,dx.\]

To conform the limits of integration, note that when \(ten=0,u=1+2(0)=1,\) and when \(10=1,u=1+two(1)=3.\) Then

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

Evaluating this expression, we go

\[ \dfrac{one}{6}∫^3_1u^v\,du=(\dfrac{1}{half dozen})(\dfrac{u^six}{6})|^3_1=\dfrac{1}{36}[(iii)^6−(one)^6]=\dfrac{182}{9}.\]

Exercise \(\PageIndex{five}\)

Use substitution to evaluate the definite integral \[ ∫^0_{−1}y(2y^two−three)^5\,dy.\]

Hint

Use the steps from Example to solve the problem.

Answer

\(\dfrac{91}{3}\)

Example \(\PageIndex{6}\): Using Commutation with an Exponential Function

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

Solution

Let \(u=4x^3+3.\) Then, \(du=8x\,dx.\) To arrange the limits of integration, we note that when \(x=0,u=3\), and when \(x=ane,u=seven\). And then our substitution gives

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

Do \(\PageIndex{6}\)

Apply substitution to evaluate \[ ∫^1_0x^2cos(\dfrac{π}{2}ten^3)\,dx.\]

Hint

Use the process from Example to solve the trouble.

Answer

\(\dfrac{two}{3π}≈0.2122\)

Exchange may exist simply 1 of the techniques needed to evaluate a definite integral. All of the properties and rules of integration utilise independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we tin can apply substitution. Likewise, nosotros accept the option of replacing the original expression for u later on nosotros find the antiderivative, which means that we practice not take to change the limits of integration. These two approaches are shown in Example.

Example \(\PageIndex{7}\): Using Exchange to Evaluate a Trigonometric Integral

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

Solution

Let us first employ a trigonometric identity to rewrite the integral. The trig identity \(\cos^2θ=\dfrac{1+\cos 2θ}{two}\) allows united states of america to rewrite the integral as

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

And so,

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

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

We tin can evaluate the starting time integral as it is, simply we need to make a commutation to evaluate the second integral. Let \(u=2θ.\) Then, \(du=ii\,dθ,\) or \(\dfrac{1}{2}\,du=dθ\). Besides, when \(θ=0,u=0,\) and when \(θ=π/ii,u=π.\) Expressing the second integral in terms of \(u\), we take

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

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

\(=(\dfrac{π}{4}−0)+(0−0)=\dfrac{π}{four}\)

Instance \( \PageIndex{8}\): Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral

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

Solution

We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, so evaluate the definite integral. We have

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

Example \( \PageIndex{9}\): Evaluating a Definite Integral

Evaluate the definite integral \( ∫^{\sqrt{iii}}_{\sqrt{iii}/3}\dfrac{dx}{1+x^ii}\).

Solution

Use the formula for the inverse tangent. Nosotros have

\[ ∫^{\sqrt{3}}_{\sqrt{3}/three}\dfrac{dx}{1+x^ii}=tan^{−ane}x∣^{\sqrt{three}}_{\sqrt{three}/3} =[tan^{−1}(\sqrt{3})]−[tan^{−one}(\dfrac{\sqrt{3}}{3})]=\dfrac{π}{6}.\]

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

\[ \dfrac{π}{viii}\]

As mentioned at the get-go of this section, exponential functions are used in many existent-life applications. The number e is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth charge per unit, the integral represents the total modify or the full growth. Let's await at an example in which integration of an exponential function solves a common business organisation awarding.

A price–demand function tells u.s. the relationship betwixt the quantity of a production demanded and the price of the product. In general, price decreases every bit quantity demanded increases. The marginal price–demand function is the derivative of the price–demand function and information technology tells us how fast the cost changes at a given level of production. These functions are used in business to determine the price–elasticity of need, and to assist companies determine whether changing production levels would exist profitable.

Example \(\PageIndex{four}\): Finding a Price–Demand Equation

Find the toll–demand equation for a particular brand of toothpaste at a supermarket chain when the need is 50 tubes per week at $2.35 per tube, given that the marginal price—need part, \(p′(10),\) for x number of tubes per week, is given as

\[p'(x)=−0.015e^{−0.01x}.\]

If the supermarket chain sells 100 tubes per week, what price should information technology prepare?

Solution

To find the price–need equation, integrate the marginal price–demand function. First find the antiderivative, and then await at the particulars. Thus,

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

Using substitution, permit \(u=−0.01x\) and \(du=−0.01dx\). Then, divide 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 side by side step is to solve for C. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means

\[p(50)=one.5e^{−0.01(50)}+C=2.35.\]

Now, merely solve for C:

\[C=2.35−1.5e^{−0.5}=ii.35−0.91=1.44.\]

Thus,

\[p(x)=1.5e^{−0.01x}+1.44.\]

If the supermarket sells 100 tubes of toothpaste per calendar week, the toll would be

\[p(100)=1.5e−0.01(100)+1.44=1.5e−1+i.44≈1.99.\]

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

Instance \(\PageIndex{5}\): Evaluating a Definite Integral Involving an Exponential Role

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

Solution

Again, substitution is the method to use. Let \(u=i−10,\) so \(du=−1dx\) or \(−du=dx\). Then \(∫eastward^{one−x}dx=−∫due east^udu.\) Next, modify the limits of integration. Using the equation \(u=1−x\), we accept

\[u=1−(ane)=0\]

\[u=1−(2)=−1.\]

The integral then becomes

\[∫^2_1e^{1−10}\,dx=−∫^{−1}_0e^u\,du=∫^0_{−1}eastward^u\,du=eu|^0_{−ane}=e^0−(e^{−one})=−e^{−1}+1.\]

See Figure.

Figure \(\PageIndex{2}\): The indicated expanse can be calculated by evaluating a definite integral using substitution.

Do \(\PageIndex{four}\)

Evaluate \(∫^2_0e^{2x}dx.\)

Hint

Let \(u=2x.\)

Answer

\(\dfrac{1}{2}∫^4_0e^udu=\dfrac{1}{ii}(east^4−1)\)

Example \(\PageIndex{6}\): Growth of Bacteria in a Culture

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

Solution

We have

\[Q(t)=∫iii^tdt=\dfrac{3^t}{\ln 3}+C.\]

And then, at \(t=0\) we have \(Q(0)=10=\dfrac{i}{\ln iii}+C,\) so \(C≈ix.090\) and we get

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

At time \(t=2\), we have

\[Q(2)=\dfrac{3^2}{\ln iii}+9.090\]

\[=17.282.\]

After 2 hours, there are 17,282 bacteria in the dish.

Do \(\PageIndex{5}\)

From Case, suppose the bacteria abound at a rate of \(q(t)=ii^t\). Assume the culture still starts with x,000 leaner. Find \(Q(t)\). How many bacteria are in the dish afterward three hours?

Hint

Utilise the process from Example to solve the problem

Answer

\(Q(t)=\dfrac{2^t}{\ln 2}+eight.557.\) There 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 charge per unit of \(grand(t)=2e^{0.02t}\), in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population later 10 days?

Solution

Allow \(G(t)\) represent the number of flies in the population at fourth dimension t. Applying the net change theorem, nosotros have

\(Chiliad(10)=G(0)+∫^{ten}_02e^{0.02t}dt\)

\(=100+[\dfrac{two}{0.02}e^{0.02t}]∣^{10}_0\)

\(=100+[100e^{0.02t}]∣^{10}_0\)

\(=100+100e^{0.2}−100\)

\(≈122.\)

There are 122 flies in the population later on 10 days.

Do \(\PageIndex{vi}\)

Suppose the rate of growth of the wing population is given by \(1000(t)=e^{0.01t},\) and the initial fly population is 100 flies. How many flies are in the population after fifteen days?

Hint

Use the process from Example to solve the problem.

Respond

There are 116 flies.

Case \(\PageIndex{8}\): Evaluating a Definite Integral Using Commutation

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{e^{1/x}}{x^2}\,dx.\]

Solution

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e equally a power of 10, then bring the \(x^2\) in the denominator up to the numerator using a negative exponent. Nosotros have

\[∫^2_1\dfrac{due east^{1/ten}}{x^2}\,dx=∫^2_1e^{x^{−1}}x^{−2}\,dx.\]

Let \(u=x^{−ane},\) the exponent on \(e\). And so

\[du=−x^{−two}\,dx\]

\[−du=x^{−ii}\,dx.\]

Bringing the negative sign outside the integral sign, the problem at present reads

\[−∫due east^u\,du.\]

Next, change the limits of integration:

\[u=(i)^{−one}=1\]

\[u=(ii)^{−ane}=\dfrac{1}{ii}.\]

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

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

Practise \(\PageIndex{7}\)

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{1}{x^3}e^{4x^{−2}}dx.\]

Hint

Allow \(u=4x^{−ii}.\)

Answer

\[∫^2_1\dfrac{i}{x^three}e^{4x^{−2}}dx=\dfrac{1}{8}[e^iv−e]\].

Example is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before nosotros can move forwards. Finding the right grade of the integrand is usually the primal to a smooth integration.

Instance \(\PageIndex{12}\): Evaluating a Definite Integral

Notice the definite integral of \[∫^{π/2}_0\dfrac{\sin 10}{ane+\cos ten}dx.\]

Solution

Nosotros need exchange to evaluate this problem. Let \(u=1+\cos 10\) so \(du=−\sin x\,dx.\) Rewrite the integral in terms of u, irresolute the limits of integration likewise. Thus,

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

\[u=1+cos(\dfrac{π}{two})=i.\]

So

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

Key Concepts

• Substitution is a technique that simplifies the integration of functions that are the event of a chain-dominion derivative. The term 'substitution' refers to changing variables or substituting the variable u and du for advisable expressions in the integrand.
• When using substitution for a definite integral, nosotros also have to modify the limits of integration.

Cardinal Equations

• Substitution with Definite Integrals

\(∫^b_af(g(x))g'(x)dx=∫^{chiliad(b)}_{k(a)}f(u)du\)

Glossary

change of variables

the substitution of a variable, such equally u, for an expression in the integrand

integration past substitution

a technique for integration that allows integration of functions that are the result of a chain-dominion 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 free 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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