I have an the integral
$$\int_{-\infty}^{6} xe^{\frac{x}{2}}\; dx$$
I know that this integral is convergent but I can not find how to evaluate its' convergence other than finding the limit of the function as it goes to negative infinity. Some guidance on this question would be super helpful!
Thank you in advance!
On
Let's look at the following integral: $$ \int_b^6 x e^{x/2} dx $$
The primitive for the integrand is known, and equals $ 2(x - 2)e^{x/2} $. Therefore: $$ \int_b^6 x e^{x/2} dx = \left[ 2(x - 2)e^{x/2} \right]_b^6 = 8e^3 - 2(b - 2) e^{b/2}$$
Now we try to evaluate the convergente of the improper integral:
$$ \int_{-\infty}^6 x e^{x/2} dx = \lim_{b \to -\infty}\int_b^6 x e^{x/2} dx = \lim_{b \to -\infty} 8e^3 - 2(b - 2) e^{b/2} = \lim_{b \to \infty} 8e^3 + \dfrac{2(b - 2)}{e^{b/2}}$$
A simple L'Hôpital on the limit will show that $ \int_{-\infty}^6 x e^{x/2} dx = 8e^3$.
You use the definition and the integration by parts. $$\begin{gathered} \int\limits_{ - \infty }^6 {x{e^{\frac{x} {2}}}dx} = \mathop {\lim }\limits_{t \to - \infty } \int\limits_t^6 {x{e^{\frac{x} {2}}}dx} = 2\mathop {\lim }\limits_{t \to - \infty } \int\limits_t^6 {xd\left( {{e^{\frac{x} {2}}}} \right)} = 2\mathop {\lim }\limits_{t \to - \infty } \left( {\left. {x{e^{\frac{x} {2}}}} \right|_{x = t}^{x = 6} - \int\limits_t^6 {{e^{\frac{x} {2}}}dx} } \right) \hfill \\ = 2\mathop {\lim }\limits_{t \to - \infty } \left( {6{e^3} - t{e^{\frac{t} {2}}} - 2\left. {{e^{\frac{x} {2}}}} \right|_t^6} \right) = 2\mathop {\lim }\limits_{t \to - \infty } \left( {6{e^3} - t{e^{\frac{t} {2}}} - 2{e^3} + 2{e^{\frac{t} {2}}}} \right) = 8{e^3} \hfill \\ \end{gathered} $$