How do you find the integral from 0 to 2 of #xe^(2x) dx#?

Question

Paisley Johnson · Accepted Answer

#\int_0^2xe^{2x}dx=3/4e^4+1/4# Make use of integration by parts #\int u dv=uv-int vdu# Let #u=x, \implies du=dx# Let #dv=e^{2x}dx, \implies v=1/2e^{2x}# Substitute #v# and #u# into the top expression #\int_0^2xe^{2x}dx=[x/2e^{2x}]_0^2 -int_0^2 1/2e^{2x}dx# #\int_0^2xe^{2x}dx=(e^4-0)-[1/4e^{2x}]_0^2# #\int_0^2xe^{2x}dx=3/4e^4+1/4#

Ezekiel Alexander · Answer

undefined To find the integral of $ xe^{2x} $ from 0 to 2, you can use integration by parts. Let $ u = x $ and $ dv = e^{2x} dx $. Then, $ du = dx $ and $ v = \frac{1}{2} e^{2x} $.

Applying the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$

We get:
$$ \int x e^{2x} \, dx = \frac{1}{2} xe^{2x} - \frac{1}{2} \int e^{2x} \, dx $$

Now, integrate $ e^{2x} $:
$$ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C $$

Substitute this back into the original equation:
$$ \int x e^{2x} \, dx = \frac{1}{2} xe^{2x} - \frac{1}{4} e^{2x} + C $$

Now, evaluate the definite integral from 0 to 2:
$$ \int_{0}^{2} x e^{2x} \, dx = \left[ \frac{1}{2} xe^{2x} - \frac{1}{4} e^{2x} \right]_{0}^{2} $$
$$ = \left( \frac{1}{2} \cdot 2 \cdot e^{4} - \frac{1}{4} \cdot e^{4} \right) - \left( \frac{1}{2} \cdot 0 \cdot e^{0} - \frac{1}{4} \cdot e^{0} \right) $$
$$ = \frac{2e^{4}}{2} - \frac{e^{4}}{4} - \frac{e^{0}}{4} $$
$$ = e^{4} - \frac{1}{4} $$

So, the integral from 0 to 2 of $ xe^{2x} $ is $ e^{4} - \frac{1}{4} $.