To solve the integral [tex]\(\int_0^2 2x e^x \, dx\)[/tex], we will use the method of integration by parts. Integration by parts is based on the formula:
[tex]\[
\int u \, dv = uv - \int v \, du
\][/tex]
where [tex]\(u\)[/tex] and [tex]\(dv\)[/tex] are parts of the integrand you choose strategically.
For our integral [tex]\(\int_0^2 2x e^x \, dx\)[/tex], let's choose:
- [tex]\(u = 2x\)[/tex]
- [tex]\(dv = e^x \, dx\)[/tex]
Next, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex]:
- [tex]\(du = 2 \, dx\)[/tex] (since the derivative of [tex]\(2x\)[/tex] is 2)
- [tex]\(v = e^x\)[/tex] (since the integral of [tex]\(e^x\)[/tex] is [tex]\(e^x\)[/tex])
Now, apply the integration by parts formula:
[tex]\[
\int_0^2 2x e^x \, dx = \left. 2x e^x \right|_0^2 - \int_0^2 2 e^x \, dx
\][/tex]
First, evaluate [tex]\(\left. 2x e^x \right|_0^2\)[/tex]:
[tex]\[
\left. 2x e^x \right|_0^2 = (2 \cdot 2 \cdot e^2) - (2 \cdot 0 \cdot e^0) = 4e^2 - 0 = 4e^2
\][/tex]
Next, solve the remaining integral [tex]\(\int_0^2 2 e^x \, dx\)[/tex]:
[tex]\[
\int_0^2 2 e^x \, dx = 2 \int_0^2 e^x \, dx
\][/tex]
Since the integral of [tex]\(e^x\)[/tex] is [tex]\(e^x\)[/tex], we get:
[tex]\[
2 \left. e^x \right|_0^2 = 2 (e^2 - e^0) = 2 (e^2 - 1) = 2e^2 - 2
\][/tex]
Putting it all together from our integration by parts result:
[tex]\[
\int_0^2 2x e^x \, dx = 4e^2 - (2e^2 - 2) = 4e^2 - 2e^2 + 2 = 2e^2 + 2
\][/tex]
Therefore, the result of the definite integral [tex]\(\int_0^2 2x e^x \, dx\)[/tex] is:
[tex]\[
2 + 2e^2
\][/tex]
