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QuestionCalculate the integral ∫02(x2+1)e−xdx\int_{0}^{2}\left(x^{2}+1\right) e^{-x} dx.

Studdy Solution

STEP 1

Assumptions1. The integral is a definite integral from0 to. . The function to be integrated is (x+1)e−x(x^{}+1) e^{-x}.
3. We will use the technique of integration by parts, which states ∫udv=uv−∫vdu\int u dv = uv - \int v du.

STEP 2

First, we need to identify the parts of the function that we will use for the integration by parts formula. We will set u=x2+1u = x^2 +1 and dv=e−xdxdv = e^{-x} dx.
u=x2+1u = x^2 +1dv=e−xdxdv = e^{-x} dx

STEP 3

Next, we need to find the derivative of uu and the integral of dvdv.
du=2xdxdu =2x dxv=−e−xv = -e^{-x}

STEP 4

Now we can apply the integration by parts formula. We will do this twice, once for each term in uu.
∫02(x2+1)e−xdx=[(x2+1)(−e−x)]02−∫02(−e−x)(2xdx)\int_{0}^{2}\left(x^{2}+1\right) e^{-x} dx = \left[(x^2 +1)(-e^{-x})\right]_{0}^{2} - \int_{0}^{2} (-e^{-x})(2x dx)

STEP 5

implify the equation.
∫02(x2+1)e−xdx=[−x2e−x−e−x]02+2∫02xe−xdx\int_{0}^{2}\left(x^{2}+1\right) e^{-x} dx = \left[-x^2 e^{-x} - e^{-x}\right]_{0}^{2} +2 \int_{0}^{2} x e^{-x} dx

STEP 6

We need to use integration by parts again for the remaining integral. Set u=xu = x and dv=e−xdxdv = e^{-x} dx.
u=xu = xdv=e−xdxdv = e^{-x} dx

STEP 7

Find the derivative of uu and the integral of dvdv.
du=dxdu = dxv=−e−xv = -e^{-x}

STEP 8

Apply the integration by parts formula to the remaining integral.
2∫02xe−xdx=2[(x)(−e−x)]02−2∫02(−e−x)dx2 \int_{0}^{2} x e^{-x} dx =2 \left[(x)(-e^{-x})\right]_{0}^{2} -2 \int_{0}^{2} (-e^{-x}) dx

STEP 9

implify the equation.
2∫2xe−xdx=−2xe−x∣2+2∫2e−xdx2 \int_{}^{2} x e^{-x} dx = -2x e^{-x} \bigg|_{}^{2} +2 \int_{}^{2} e^{-x} dx

STEP 10

The remaining integral is a standard integral. Evaluate it.
2∫02e−xdx=2[−e−x]022 \int_{0}^{2} e^{-x} dx =2 \left[-e^{-x}\right]_{0}^{2}

STEP 11

implify the equation.
∫0e−xdx=e−− \int_{0}^{} e^{-x} dx =e^{-} -

STEP 12

Substitute the results from steps5,9, and11 back into the original equation.
∫02(x2+)e−xdx=[−x2e−x−e−x]02−2xe−x∣02+2e−2−2\int_{0}^{2}\left(x^{2}+\right) e^{-x} dx = \left[-x^2 e^{-x} - e^{-x}\right]_{0}^{2} -2x e^{-x} \bigg|_{0}^{2} +2e^{-2} -2

STEP 13

Evaluate the equation at the limits of0 and2.
∫02(x2+)e−xdx=−e−2−e−2+2e−2−2+0−=−3e−2−3\int_{0}^{2}\left(x^{2}+\right) e^{-x} dx = -e^{-2} - e^{-2} +2e^{-2} -2 +0 - = -3e^{-2} -3The solution to the integral ∫02(x2+)e−xdx\int_{0}^{2}\left(x^{2}+\right) e^{-x} dx is −3e−2−3-3e^{-2} -3.

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