1. ## cal 2

intergral x(x^2-9)^3 dx
intergral with limits 3/2 and 0 xsin(x^2)dx give answer in decimal number.
intergral with limits 4 and -2 x^2(x^3 + 8)^2dx. give answer in decimal number.

2. Originally Posted by lay137
intergral x(x^2-9)^3 dx
$\displaystyle \int x(x^2-9)^3 dx$
Use $\displaystyle u=x^2-9$.
Then you get,
$\displaystyle \frac{1}{2} \int u^3 du$
$\displaystyle \frac{1}{2} \cdot \frac{1}{4}u^4 +C$
$\displaystyle \frac{1}{8}u^4+C$
$\displaystyle \frac{1}{8}(x^2-9)^4+C$

intergral with limits 3/2 and 0 xsin(x^2)dx give answer in decimal number.
$\displaystyle \int_0^{3/2} x\sin x^2 dx$
Let $\displaystyle u=x^2$ then,
$\displaystyle \frac{1}{2}\int_0^{9/4} \sin u du$
$\displaystyle \frac{1}{2} \left( -\cos u \big|_0^{9/4} \right)$
You can evaluate that.
intergral with limits 4 and -2 x^2(x^3 + 8)^2dx. give answer in decimal number.
$\displaystyle \int_{-2}^4 -2 x^2(x^3+8)^2 dx$
Let $\displaystyle u=x^3+8$ then,
$\displaystyle \frac{1}{3}\int_0^72 -2 u^2 du$
$\displaystyle \frac{1}{3} \left( \frac{-2}{3} u^3 \big|_0^{72} \right)$