Question

If $\int f\left(x\right)dx=F\left(x\right)$, then $\int x^{3}f\left(x^2\right)dx$ equals to

• A. $\frac{1}{2}\left[x^2\right(F\left(x\right){)}^2-\int \left(F\right(x\left){)}^{2}dx\right]$
• B. $\frac{1}{2}\left[x^{2}F\right(x^2)-\int F(x^2\left)d\right(x^2\left)\right]$
• C. $\frac{1}{2}\left[x^{2}F\right(x)-\frac{1}{2}\int (F\left(x\right){)}^{2}dx]$
• D. $\frac{1}{2}\left[x^{2}F\right(x^2)+\int F(x^2\left)d\right(x^2\left)\right]$

Answer 1

Answer: (B)

$\frac{1}{2}\left[x^{2}F\right(x^2)-\int F(x^2\left)d\right(x^2\left)\right]$

Given $\int f\left(x\right)dx=F\left(x\right)$

Let $I=\int x^{3}f\left(x^2\right)dx$
Put $x^2=t$
$\Rightarrow xdx=\frac{dt}{2}$
$\Rightarrow I=\frac{1}{2}\int tf\left(t\right)dt$
$=\frac{1}{2}[t\int f(t)dt-\int (\int f\left(t\right)dt)dt]$ $\left[\text{Integration by parts}\int (u\cdot v)dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx\right]$
$\Rightarrow I=\frac{1}{2}\left[tF\right(t)-\int F(t\left)dt\right]$
$\Rightarrow I=\frac{1}{2}\left[x^{2}F\right(x^2)-\int F(x^2\left)d\right(x^2\left)\right]$