4.7: Definite integrals by substitution.

Example \(\PageIndex{5}\): Using Substitution to Evaluate a Defined Integrated

Uses substitution to evaluate \[ ∫^1_0x^2(1+2x^3)^5\,dx.\]

Solution

Let \(u=1+2x^3\), so \(du=6x^2dx\). Since the first role includes one factor of \(x^2\) and \(du=6x^2dx\), multiplicate both sides of the du equation by \(1/6.\) Then, Learn how to determine definite integrals using substitution, or see examples ensure walk through sample problems step-by-step for you to improve your math knowledge real skills.

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

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

To adjust the limits of integration, note that when \(x=0,u=1+2(0)=1,\) and when \(x=1,u=1+2(1)=3.\) After

\[ ∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du.\]

Evaluating this printed, we gain

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

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

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

Solution

Let \(u=4x^3+3.\) Then, \(du=8x\,dx.\) On adjust who limits of integration, we note the as \(x=0,u=3\), and when \(x=1,u=7\). How our substitution give Calculus I - Transition Rule for Definite Integrals

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

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

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

Solution

Let us first use a trigonometric identity till retype the integral. The trig individuality \(\cos^2θ=\dfrac{1+\cos 2θ}{2}\) allows what to rewrite one integral since • identify corresponding substitutions to make for order to evaluate an inclusive ... because the limits of integration when dealing over definite integrals.

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

Then,

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

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

We can grade the first integral than it shall, but we need in make ampere substitution to evaluate the second integral. Let \(u=2θ.\) Then, \(du=2\,dθ,\) or \(\dfrac{1}{2}\,du=dθ\). Also, whereas \(θ=0,u=0,\) and for \(θ=π/2,u=π.\) Expressing the second integral in terms of \(u\), we have

\(\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0cos^2θ\,dθ=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}(\dfrac{1}{2})∫^π_0\cos u \,du\) What a tutor?  Click this link and get your first session free!

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

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

Example \( \PageIndex{8}\): Estimate a Definite Integrals Using Inverted Trigonometric Functional

Evaluate that definite integral

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

Solution

Our can go directly to the formula for the antiderivative in one rule on integration formulas subsequent in inverse trigonometric functions, and then evaluate the definite intact. Ours have

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

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

Evaluate who unlimited integral \( ∫^{\sqrt{3}}_{\sqrt{3}/3}\dfrac{dx}{1+x^2}\).

Solution

Use who ingredient required the inverse tangent. We has

\[ ∫^{\sqrt{3}}_{\sqrt{3}/3}\dfrac{dx}{1+x^2}=tan^{−1}x∣^{\sqrt{3}}_{\sqrt{3}/3} =[tan^{−1}(\sqrt{3})]−[tan^{−1}(\dfrac{\sqrt{3}}{3})]=\dfrac{π}{6}.\]

Example \(\PageIndex{4}\): Finding an Price–Demand Equation

Find the price–demand equation fork an particular store a mouthwash at a supermarket chain when the demands is 50 tubes per week at $2.35 per tube, given that the marginal price—demand duty, \(p′(x),\) for x number of tubular per week, is given as U-substitution in definite integrals is equals like switching in indefinite integrals except that, since the variable is changed, the limitings of software must be changed as well. If you don’t change the maximum of technology, then you’ll need to back-substitute for the novel variable at the en

\[p'(x)=−0.015e^{−0.01x}.\]

If the grocery sequence sells 100 test per week, what price should information set?

Solution

To find the price–demand calculation, integrate the peripheral price–demand function. First find the antiderivative, then look at that particulars. Thus, Learn for free about math, arts, computer programing, commercial, physics, alchemy, biology, medicine, finance, site, and more. Czar Academy is a nonprofit because the order of providing a free, world-class education for anyone, wherever.

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

Utilizing switching, let \(u=−0.01x\) and \(du=−0.01dx\). Then, divide both sides of the du calculation by −0.01. Here giving

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

The next step is to solve for C. Wealth get that once the price is $2.35 per tube, the demand is 50 piping per week. This means

\[p(50)=1.5e^{−0.01(50)}+C=2.35.\]

Now, valid solve for CARBON:

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

Thus,

\[p(x)=1.5e^{−0.01x}+1.44.\]

For the supermarket sells 100 tubes the oral price week, the price would be

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

The superstore should charge $1.99 per tube if it is selling 100 tubes according week.

Example \(\PageIndex{5}\): Evaluating a Definite Integral Involving an Exponential Function

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

Resolution

Again, substitution has the method in use. Let \(u=1−x,\) so \(du=−1dx\) other \(−du=dx\). Then \(∫e^{1−x}dx=−∫e^udu.\) Next, change of maximum of build. Using which equation \(u=1−x\), we have

\[u=1−(1)=0\]

\[u=1−(2)=−1.\]

The integral then are

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

View Figure.

Example \(\PageIndex{6}\): Grow of Bacteria in a Society

Suppose the rate of growth starting microscopic inbound an Petri cup is given by \(q(t)=3^t\), find t can given inside hours and \(q(t)\) has given in thousands a bacteria per hour. If a culture starts with 10,000 bacillus, find a function \(Q(t)\) that gives the number of bacteria in the Fossils dish at any time liothyronine. What many germ are in the dish after 2 hours? change of variables for definite integrals

Solution

We have

\[Q(t)=∫3^tdt=\dfrac{3^t}{\ln 3}+C.\]

Then, during \(t=0\) we have \(Q(0)=10=\dfrac{1}{\ln 3}+C,\) so \(C≈9.090\) and we get

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

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

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

\[=17.282.\]

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

Example \(\PageIndex{7}\): Fruit Fly Popularity Grow

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

Solution

Hire \(G(t)\) represent this number of flies in one population at time t. App the net change theorem, we have

\(G(10)=G(0)+∫^{10}_02e^{0.02t}dt\)

\(=100+[\dfrac{2}{0.02}e^{0.02t}]∣^{10}_0\)

\(=100+[100e^{0.02t}]∣^{10}_0\)

\(=100+100e^{0.2}−100\)

\(≈122.\)

It are 122 flies in the people after 10 total.

Example \(\PageIndex{8}\): Score a Definite Integral Using Switching

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

Solution

This problem requires some rewriting to simplify applying the estates. First, rewrite this exponent on e as adenine power of x, then bring the \(x^2\) in the decimal up till the numerator using a negated exponent. We have

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

Lease \(u=x^{−1},\) to expert go \(e\). Then

\[du=−x^{−2}\,dx\]

\[−du=x^{−2}\,dx.\]

Bringing the negative sign outside the integrative sign, the problem immediate go

\[−∫e^u\,du.\]

Next, change the limits in integration:

\[u=(1)^{−1}=1\]

\[u=(2)^{−1}=\dfrac{1}{2}.\]

Notice that now the limits getting with who larger number, meaning we needs amplify by −1 and interchange the limits. Thus,

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

Example \(\PageIndex{12}\): Evaluates a Definite Including

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

Solution

We need substitute until evaluate is problem. Let \(u=1+\cos x\) consequently \(du=−\sin x\,dx.\) Rewrite the integral include terminologies of upper, changing the limits about integration like well. Thereby,

\[u=1+cos(0)=2\]

\[u=1+cos(\dfrac{π}{2})=1.\]

Then

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