# Solve the following integral \int_{0}^{4}3x(4-x)dx=32 \int_{0}^{4}x(x-4)dx

Solve the following integral
${\int }_{0}^{4}3x\left(4-x\right)dx=32$
${\int }_{0}^{4}x\left(x-4\right)dx$
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wheezym
Step 1
${\int }_{0}^{4}3x\left(4-x\right)dx=32$
We have to evaluate.
${\int }_{0}^{4}x\left(x-4\right)dx$
Step 2
We have, ${\int }_{0}^{4}3x\left(4-x\right)dx=32$...(1)
and ${\int }_{0}^{4}x\left(x-4\right)dx=$?
Now,
$⇒{\int }_{0}^{4}3x\left(4-x\right)dx=32$
$⇒3{\int }_{0}^{4}x\left(4-x\right)dx=32$
$⇒{\int }_{0}^{4}x\left(4-x\right)dx=\frac{32}{3}$
$⇒{\int }_{0}^{4}x\left(-1\right)\left(x-4\right)dx=\frac{32}{3}$
$⇒-{\int }_{0}^{4}x\left(x-4\right)dx=\frac{32}{3}$
$⇒{\int }_{0}^{4}x\left(x-4\right)dx=-\frac{32}{3}$
Hence, ${\int }_{0}^{4}x\left(x-4\right)dx=-\frac{32}{3}$
Jeffrey Jordon