Phoebe

2020-11-12

Solve the following integral

${\int}_{0}^{4}3x(4-x)dx=32$

${\int}_{0}^{4}x(x-4)dx$

wheezym

Skilled2020-11-13Added 103 answers

Step 1

${\int}_{0}^{4}3x(4-x)dx=32$

We have to evaluate.

${\int}_{0}^{4}x(x-4)dx$

Step 2

We have,${\int}_{0}^{4}3x(4-x)dx=32$ ...(1)

and${\int}_{0}^{4}x(x-4)dx=$ ?

Now,

$\Rightarrow {\int}_{0}^{4}3x(4-x)dx=32$

$\Rightarrow 3{\int}_{0}^{4}x(4-x)dx=32$

$\Rightarrow {\int}_{0}^{4}x(4-x)dx=\frac{32}{3}$

$\Rightarrow {\int}_{0}^{4}x(-1)(x-4)dx=\frac{32}{3}$

$\Rightarrow -{\int}_{0}^{4}x(x-4)dx=\frac{32}{3}$

$\Rightarrow {\int}_{0}^{4}x(x-4)dx=-\frac{32}{3}$

Hence,$\int}_{0}^{4}x(x-4)dx=-\frac{32}{3$

We have to evaluate.

Step 2

We have,

and

Now,

Hence,

Jeffrey Jordon

Expert2021-11-05Added 2575 answers

Answer is given below (on video)