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

Question
Applications of integrals
Solve the following integral
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{3}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}$$
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}$$

2020-11-13
Step 1
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{3}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}$$
We have to evaluate.
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}$$
Step 2
We have, $$\displaystyle{\int_{{{0}}}^{{{4}}}}{3}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}$$...(1)
and $$\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}=$$?
Now,
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{4}}}}{3}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}$$
$$\displaystyle\Rightarrow{3}{\int_{{{0}}}^{{{4}}}}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}$$
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{4}}}}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={\frac{{{32}}}{{{3}}}}$$
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{4}}}}{x}{\left(-{1}\right)}{\left({x}-{4}\right)}{\left.{d}{x}\right.}={\frac{{{32}}}{{{3}}}}$$
$$\displaystyle\Rightarrow-{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}={\frac{{{32}}}{{{3}}}}$$
$$\displaystyle\Rightarrow{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}=-{\frac{{{32}}}{{{3}}}}$$
Hence, $$\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}=-{\frac{{{32}}}{{{3}}}}$$

### Relevant Questions

Evaluate the following integral.
$$\displaystyle{\int_{{{1}}}^{{{32}}}}{\left({x}^{{-{\frac{{{6}}}{{{5}}}}}}\right)}{\left.{d}{x}\right.}$$
Solve the following integral. $$\displaystyle{\int_{{{1}}}^{{{e}}}}{x}^{{{2}}}{\ln{{x}}}{\left.{d}{x}\right.}$$
Evaluate the following definite integral.
$$\displaystyle{\int_{{{0}}}^{{{\frac{{\pi}}{{{2}}}}}}}{x}{\cos{{2}}}{x}{\left.{d}{x}\right.}$$
Calculate the iterated integral.
$$\displaystyle{\int_{{{0}}}^{{{4}}}}{\int_{{{0}}}^{{{12}}}}{2}{e}^{{{x}+{3}{y}}}{\left.{d}{x}\right.}\ {\left.{d}{y}\right.}$$
Evaluate the following definite integrals
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({x}^{{{4}}}+{7}{e}^{{{x}}}-{3}\right)}{\left.{d}{x}\right.}$$
Which of the following integrals are improper integrals?
1.$$\displaystyle{\int_{{{0}}}^{{{3}}}}{\left({3}-{x}\right)}^{{{2}}}{\left\lbrace{3}\right\rbrace}{\left.{d}{x}\right.}$$
2.$$\displaystyle{\int_{{{1}}}^{{{16}}}}{\frac{{{e}^{{\sqrt{{{x}}}}}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}$$
3.$$\displaystyle{\int_{{{1}}}^{{\propto}}}{\frac{{{3}}}{{\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}}}}{\left.{d}{x}\right.}$$
4.$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{3}{\left({x}+{1}\right)}^{{-{1}}}{\left.{d}{x}\right.}$$
a) 1 only
b)1 and 2
c)3 only
d)2 and 3
e)1,3 and 4
f)All of the integrals are improper
$$\displaystyle{\int_{{{0}}}^{{\pi}}}{x}{\sin{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{-{2}}}^{{{2}}}}{\left({3}{x}^{{{4}}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{-{{2}}}}^{{2}}}{\left({3}{x}^{{4}}-{2}{x}+{1}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{x}{e}^{{{\left(-{x}^{{{2}}}+{2}\right)}}}{\left.{d}{x}\right.}$$