Solved on Feb 05, 2024

Evaluate the definite integral of $\frac{4y}{e^{2y}}$ from 0 to 1.

STEP 1

Assumptions
1. We are given the integral $\int_{0}^{1} \frac{4 y}{e^{2 y}} d y$ .
2. We will use the method of integration by substitution to solve the integral.
3. The bounds of the integral are from 0 to 1.

STEP 2

Choose a substitution that will simplify the integral. Let $u = 2y$ , which implies that $du = 2 dy$ or $dy = \frac{du}{2}$ .

STEP 3

Change the limits of integration according to the substitution. When $y = 0$ , $u = 2(0) = 0$ . When $y = 1$ , $u = 2(1) = 2$ .

STEP 4

Rewrite the integral in terms of $u$ using the substitution.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = \int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2}$

STEP 5

Simplify the integral by combining the constants and the $u$ term.
$\int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2}$

STEP 6

Further simplify the integral by canceling out the 2's.
$\int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{u}{e^{u}} du$

STEP 7

Now integrate the function $\frac{u}{e^{u}}$ with respect to $u$ from 0 to 2. This is an integration by parts problem, where we let $dv = e^{-u} du$ and $u = u$ .

STEP 8

Using integration by parts, where $\int u dv = uv - \int v du$ , we let $u = u$ and $dv = e^{-u} du$ .

STEP 9

Differentiate $u$ to get $du$ and integrate $dv$ to get $v$ .
$du = du$ $v = \int e^{-u} du = -e^{-u}$

STEP 10

Apply the integration by parts formula.
$\int u dv = uv - \int v du$

STEP 11

Substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula.
$\int_{0}^{2} \frac{u}{e^{u}} du = \left. -u e^{-u} \right|_{0}^{2} - \int_{0}^{2} -e^{-u} du$

STEP 12

Simplify the integral and evaluate the remaining integral.
$\int_{0}^{2} \frac{u}{e^{u}} du = -2 e^{-2} - (-e^{-u})\Big|_{0}^{2}$

STEP 13

Evaluate the antiderivative at the upper and lower limits.
$-2 e^{-2} - (-e^{-u})\Big|_{0}^{2} = -2 e^{-2} - (-e^{-2} + e^{0})$

STEP 14

Simplify the expression by combining like terms and using the fact that $e^{0} = 1$ .
$-2 e^{-2} - (-e^{-2} + e^{0}) = -2 e^{-2} + e^{-2} - 1$

STEP 15

Combine the terms with $e^{-2}$ .
$-2 e^{-2} + e^{-2} - 1 = -e^{-2} - 1$

STEP 16

The final answer is the negative of the sum of $e^{-2}$ and 1.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = -e^{-2} - 1$

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Evaluate the definite integral of $\frac{4y}{e^{2y}}$ from 0 to 1.

STEP 1

Assumptions
1. We are given the integral $\int_{0}^{1} \frac{4 y}{e^{2 y}} d y$ .
2. We will use the method of integration by substitution to solve the integral.
3. The bounds of the integral are from 0 to 1.

STEP 2

Choose a substitution that will simplify the integral. Let $u = 2y$ , which implies that $du = 2 dy$ or $dy = \frac{du}{2}$ .

STEP 3

Change the limits of integration according to the substitution. When $y = 0$ , $u = 2(0) = 0$ . When $y = 1$ , $u = 2(1) = 2$ .

STEP 4

Rewrite the integral in terms of $u$ using the substitution.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = \int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2}$

STEP 5

Simplify the integral by combining the constants and the $u$ term.
$\int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2}$

STEP 6

Further simplify the integral by canceling out the 2's.
$\int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{u}{e^{u}} du$

STEP 7

Now integrate the function $\frac{u}{e^{u}}$ with respect to $u$ from 0 to 2. This is an integration by parts problem, where we let $dv = e^{-u} du$ and $u = u$ .

STEP 8

Using integration by parts, where $\int u dv = uv - \int v du$ , we let $u = u$ and $dv = e^{-u} du$ .

STEP 9

Differentiate $u$ to get $du$ and integrate $dv$ to get $v$ .
$du = du$ $v = \int e^{-u} du = -e^{-u}$

STEP 10

Apply the integration by parts formula.
$\int u dv = uv - \int v du$

STEP 11

Substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula.
$\int_{0}^{2} \frac{u}{e^{u}} du = \left. -u e^{-u} \right|_{0}^{2} - \int_{0}^{2} -e^{-u} du$

STEP 12

Simplify the integral and evaluate the remaining integral.
$\int_{0}^{2} \frac{u}{e^{u}} du = -2 e^{-2} - (-e^{-u})\Big|_{0}^{2}$

STEP 13

Evaluate the antiderivative at the upper and lower limits.
$-2 e^{-2} - (-e^{-u})\Big|_{0}^{2} = -2 e^{-2} - (-e^{-2} + e^{0})$

STEP 14

Simplify the expression by combining like terms and using the fact that $e^{0} = 1$ .
$-2 e^{-2} - (-e^{-2} + e^{0}) = -2 e^{-2} + e^{-2} - 1$

STEP 15

Combine the terms with $e^{-2}$ .
$-2 e^{-2} + e^{-2} - 1 = -e^{-2} - 1$

STEP 16

The final answer is the negative of the sum of $e^{-2}$ and 1.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = -e^{-2} - 1$

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Evaluate the definite integral of 4ye2y\frac{4y}{e^{2y}}e2y4y​ from 0 to 1.

STEP 1

Assumptions1. We are given the integral ∫014ye2ydy\int_{0}^{1} \frac{4 y}{e^{2 y}} d y∫01​e2y4y​dy.2. We will use the method of integration by substitution to solve the integral.3. The bounds of the integral are from 0 to 1.

STEP 2

Choose a substitution that will simplify the integral. Let u=2yu = 2yu=2y, which implies that du=2dydu = 2 dydu=2dy or dy=du2dy = \frac{du}{2}dy=2du​.

STEP 3

Change the limits of integration according to the substitution. When y=0y = 0y=0, u=2(0)=0u = 2(0) = 0u=2(0)=0. When y=1y = 1y=1, u=2(1)=2u = 2(1) = 2u=2(1)=2.

STEP 4

Rewrite the integral in terms of uuu using the substitution.∫014ye2ydy=∫024u2eu⋅du2\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = \int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2}∫01​e2y4y​dy=∫02​eu42u​​⋅2du​

STEP 5

Simplify the integral by combining the constants and the uuu term.∫024u2eu⋅du2=∫022ueu⋅du2\int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2}∫02​eu42u​​⋅2du​=∫02​eu2u​⋅2du​

STEP 6

Further simplify the integral by canceling out the 2's.∫022ueu⋅du2=∫02ueudu\int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{u}{e^{u}} du∫02​eu2u​⋅2du​=∫02​euu​du

STEP 7

Now integrate the function ueu\frac{u}{e^{u}}euu​ with respect to uuu from 0 to 2. This is an integration by parts problem, where we let dv=e−ududv = e^{-u} dudv=e−udu and u=uu = uu=u.

STEP 8

Using integration by parts, where ∫udv=uv−∫vdu\int u dv = uv - \int v du∫udv=uv−∫vdu, we let u=uu = uu=u and dv=e−ududv = e^{-u} dudv=e−udu.

STEP 9

Differentiate uuu to get dududu and integrate dvdvdv to get vvv.du=dudu = dudu=du v=∫e−udu=−e−uv = \int e^{-u} du = -e^{-u}v=∫e−udu=−e−u

STEP 10

Apply the integration by parts formula.∫udv=uv−∫vdu\int u dv = uv - \int v du∫udv=uv−∫vdu

STEP 11

Substitute uuu, vvv, dududu, and dvdvdv into the integration by parts formula.∫02ueudu=−ue−u∣02−∫02−e−udu\int_{0}^{2} \frac{u}{e^{u}} du = \left. -u e^{-u} \right|_{0}^{2} - \int_{0}^{2} -e^{-u} du∫02​euu​du=−ue−u∣∣​02​−∫02​−e−udu

STEP 12

Simplify the integral and evaluate the remaining integral.∫02ueudu=−2e−2−(−e−u)∣02\int_{0}^{2} \frac{u}{e^{u}} du = -2 e^{-2} - (-e^{-u})\Big|_{0}^{2}∫02​euu​du=−2e−2−(−e−u)∣∣​02​

STEP 13

Evaluate the antiderivative at the upper and lower limits.−2e−2−(−e−u)∣02=−2e−2−(−e−2+e0)-2 e^{-2} - (-e^{-u})\Big|_{0}^{2} = -2 e^{-2} - (-e^{-2} + e^{0})−2e−2−(−e−u)∣∣​02​=−2e−2−(−e−2+e0)

STEP 14

Simplify the expression by combining like terms and using the fact that e0=1e^{0} = 1e0=1.−2e−2−(−e−2+e0)=−2e−2+e−2−1-2 e^{-2} - (-e^{-2} + e^{0}) = -2 e^{-2} + e^{-2} - 1−2e−2−(−e−2+e0)=−2e−2+e−2−1

STEP 15

Combine the terms with e−2e^{-2}e−2.−2e−2+e−2−1=−e−2−1-2 e^{-2} + e^{-2} - 1 = -e^{-2} - 1−2e−2+e−2−1=−e−2−1

STEP 16

The final answer is the negative of the sum of e−2e^{-2}e−2 and 1.∫014ye2ydy=−e−2−1\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = -e^{-2} - 1∫01​e2y4y​dy=−e−2−1

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Evaluate the definite integral of $\frac{4y}{e^{2y}}$ from 0 to 1.

Assumptions
1. We are given the integral $\int_{0}^{1} \frac{4 y}{e^{2 y}} d y$ .
2. We will use the method of integration by substitution to solve the integral.
3. The bounds of the integral are from 0 to 1.

Choose a substitution that will simplify the integral. Let $u = 2y$ , which implies that $du = 2 dy$ or $dy = \frac{du}{2}$ .

Change the limits of integration according to the substitution. When $y = 0$ , $u = 2(0) = 0$ . When $y = 1$ , $u = 2(1) = 2$ .

Rewrite the integral in terms of $u$ using the substitution.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = \int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2}$

Simplify the integral by combining the constants and the $u$ term.
$\int_{0}^{2} \frac{4 \frac{u}{2}}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2}$

Further simplify the integral by canceling out the 2's.
$\int_{0}^{2} \frac{2u}{e^{u}} \cdot \frac{du}{2} = \int_{0}^{2} \frac{u}{e^{u}} du$

Now integrate the function $\frac{u}{e^{u}}$ with respect to $u$ from 0 to 2. This is an integration by parts problem, where we let $dv = e^{-u} du$ and $u = u$ .

Using integration by parts, where $\int u dv = uv - \int v du$ , we let $u = u$ and $dv = e^{-u} du$ .

Differentiate $u$ to get $du$ and integrate $dv$ to get $v$ .
$du = du$ $v = \int e^{-u} du = -e^{-u}$

Apply the integration by parts formula.
$\int u dv = uv - \int v du$

Substitute $u$ , $v$ , $du$ , and $dv$ into the integration by parts formula.
$\int_{0}^{2} \frac{u}{e^{u}} du = \left. -u e^{-u} \right|_{0}^{2} - \int_{0}^{2} -e^{-u} du$

Simplify the integral and evaluate the remaining integral.
$\int_{0}^{2} \frac{u}{e^{u}} du = -2 e^{-2} - (-e^{-u})\Big|_{0}^{2}$

Evaluate the antiderivative at the upper and lower limits.
$-2 e^{-2} - (-e^{-u})\Big|_{0}^{2} = -2 e^{-2} - (-e^{-2} + e^{0})$

Simplify the expression by combining like terms and using the fact that $e^{0} = 1$ .
$-2 e^{-2} - (-e^{-2} + e^{0}) = -2 e^{-2} + e^{-2} - 1$

Combine the terms with $e^{-2}$ .
$-2 e^{-2} + e^{-2} - 1 = -e^{-2} - 1$

The final answer is the negative of the sum of $e^{-2}$ and 1.
$\int_{0}^{1} \frac{4 y}{e^{2 y}} d y = -e^{-2} - 1$