How do you evaluate the integral #((x^2)+1) e^-x dx# from 0 to 1?

Question

Elijah Abram · Accepted Answer

# 3/e(e-2)#.

Let, #I=int(x^2+1)e^-xdx#. Using the following Rule of Integration by Parts (IBP) : IBP : #intuv'dx=uv-intvu'dx#. We take, #u=x^2+1, and, v'=e^-x#. #:. u'=2x, and, v=inte^-xdx=e^-x/-1=-e^-x#. #:. I=-(x^2+1)e^-x-int(-e^-x*2x)dx#, #rArr I=-(x^2+1)e^-x+2I_1," where, "I_1=intxe^-xdx#. We once again use IBP for #I_1#; this time, we choose, #u=x, and, v'=e^-x :. u'=1, and, v=-e^-x#. #:. I_1=-xe^-x-int(-e^-x*1)dx#, #: I_1=-xe^-x-e^-x#. Utilising #I_1# in #I#, we get, #I=-(x^2+1)e^-x+2{-xe^-x-e^-x}#, # rArr I=-(x^2+2x+3)e^-x+C#. #:. int_0^1(x^2+1)e^-xdx=-[(x^2+2x+3)e^-x]_0^1#, #=-[(1+2+3)e^-1-(0+0+3)e^-0]#. # rArr int_0^1(x^2+1)e^-xdx=-(6/e-3)=3/e(e-2)#. Enjoy Maths.!

Emma Aldridge · Answer

undefined To evaluate the integral $\int_{0}^{1} (x^2 + 1) e^{-x} \,dx$, we can use integration by parts. The formula for integration by parts is:

$$
\int u \, dv = uv - \int v \, du
$$

Let's choose $u = x^2 + 1$ and $dv = e^{-x} \, dx$. Then, we have $du = 2x \, dx$ and $v = -e^{-x}$. Applying the formula, we get:

$$
\begin{aligned}
\int_{0}^{1} (x^2 + 1) e^{-x} \,dx & = \left. -(x^2 + 1)e^{-x} \right|_{0}^{1} - \int_{0}^{1} -2xe^{-x} \,dx \
& = -(2e^{-1} - e^{-1}) - \left. 2xe^{-x} \right|_{0}^{1} + \int_{0}^{1} 2e^{-x} \,dx \
& = -e^{-1} - \left(2e^{-1} - e^{-1}\right) - \left. 2xe^{-x} \right|_{0}^{1} \
& = -3e^{-1} - \left. 2xe^{-x} \right|_{0}^{1} \
& = -3e^{-1} - (2e^{-1} - 0) \
& = -5e^{-1} \
& \approx -1.84
\end{aligned}
$$

So, the value of the integral $\int_{0}^{1} (x^2 + 1) e^{-x} \,dx$ is approximately $-1.84$.