What is &#x222B;<!-- ∫ --> 2 3 </mfrac> <mrow class="MJX-TeXAtom-ORD"> &#xFEFF

Ashley Fritz

Ashley Fritz

Answered question

2022-05-06

What is 2 3  e x x 2 d x?

Answer & Explanation

Julius Johnston

Julius Johnston

Beginner2022-05-07Added 17 answers

Simplify.

2e-xx23dx

Since 23 is constant with respect to x, move 23 out of the integral.

23e-xx2dx

Integrate by parts using the formula udv=uv-vdu, where u=x2 and dv=e-x.

23(x2(-e-x)--e-x(2x)dx)

Multiply 2 by -1.

23(x2(-e-x)--2e-xxdx)

Since -2 is constant with respect to x, move -2 out of the integral.

23(x2(-e-x)-(-2e-xxdx))

Multiply -2 by -1.

23(x2(-e-x)+2e-xxdx)

Integrate by parts using the formula udv=uv-vdu, where u=x and dv=e-x.

23(x2(-e-x)+2(x(-e-x)--e-xdx))

Since -1 is constant with respect to x, move -1 out of the integral.

23(x2(-e-x)+2(x(-e-x)--e-xdx))

Simplify.

23(x2(-e-x)+2(x(-e-x)+e-xdx))

Let u=-x. Then du=-dx, so -du=dx. Rewrite using u and du.

23(x2(-e-x)+2(x(-e-x)+-eudu))

Since -1 is constant with respect to u, move -1 out of the integral.

23(x2(-e-x)+2(x(-e-x)-eudu))

The integral of eu with respect to u is eu.

23(x2(-e-x)+2(x(-e-x)-(eu+C)))

Rewrite 23(x2(-e-x)+2(x(-e-x)-(eu+C))) as 23(-x2e-x-2xe-x-2eu)+C.

23(-x2e-x-2xe-x-2eu)+C

Replace all occurrences of u with -x.

23(-x2e-x-2xe-x-2eu)+C

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