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

Studdy Solution

STEP 1

Assumptions
1. We are given the integral ∫014ye2ydy\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=2yu = 2y, which implies that du=2dydu = 2 dy or dy=du2dy = \frac{du}{2}.

STEP 3

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

STEP 4

Rewrite the integral in terms of uu 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}

STEP 5

Simplify the integral by combining the constants and the uu 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}

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

STEP 7

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

STEP 8

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

STEP 9

Differentiate uu to get dudu and integrate dvdv to get vv.
du=dudu = du v=∫e−udu=−e−uv = \int e^{-u} du = -e^{-u}

STEP 10

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

STEP 11

Substitute uu, vv, dudu, and dvdv 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

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}

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})

STEP 14

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

STEP 15

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

STEP 16

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

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