How do you integrate #int_-1^1x(1+x)^3dx#?

Question

Christopher Allers · Accepted Answer

Do a "u" substitute

Given: #int_-1^1x(1+x)^3dx# let #u = x + 1#, then #du = dx and x = u -1# Change the limits: #a = -1 + 1# #a = 0# #b = 1 + 1# #b = 2# #int_-1^1x(1+x)^3dx = int_0^2(u-1)u^3du# #int_-1^1x(1+x)^3dx = int_0^2u^4-u^3du# #int_-1^1x(1+x)^3dx = u^5/5-u^4/4|_0^2# #int_-1^1x(1+x)^3dx = 2^5/5-2^4/4# #int_-1^1x(1+x)^3dx = 12/5#

Aurora Alvarado · Answer

undefined To integrate $\int_{-1}^{1}x(1+x)^3 \, dx$, you can use the method of integration by parts. Let $u = x$ and $dv = (1+x)^3 \, dx$. Then, $du = dx$ and $v = \frac{1}{4}(1+x)^4$. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$$
\begin{aligned}
\int x(1+x)^3 \, dx &= \frac{1}{4}x(1+x)^4 - \int \frac{1}{4}(1+x)^4 \, dx \
&= \frac{1}{4}x(1+x)^4 - \frac{1}{4} \cdot \frac{1}{5}(1+x)^5 + C,
\end{aligned}
$$

where $C$ is the constant of integration. Finally, evaluate the definite integral from $-1$ to $1$ by substituting the limits:

$$
\begin{aligned}
\int_{-1}^{1}x(1+x)^3 \, dx &= \left[\frac{1}{4}x(1+x)^4 - \frac{1}{4} \cdot \frac{1}{5}(1+x)^5\right]_{-1}^{1} \
&= \left(\frac{1}{4}(1)(2)^4 - \frac{1}{4} \cdot \frac{1}{5}(2)^5\right) - \left(\frac{1}{4}(-1)(0)^4 - \frac{1}{4} \cdot \frac{1}{5}(0)^5\right) \
&= \frac{1}{4}(2^5) - \frac{1}{4} \cdot \frac{1}{5}(2^5) \
&= \frac{31}{5}.
\end{aligned}
$$

So, the value of the integral $\int_{-1}^{1}x(1+x)^3 \, dx$ is $\frac{31}{5}$.