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

Question
Applications of integrals
asked 2020-11-12
Solve the following integral
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{3}{x}{\left({4}-{x}\right)}{\left.{d}{x}\right.}={32}\)
\(\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}\)

Answers (1)

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, \(\displaystyle{\int_{{{0}}}^{{{4}}}}{x}{\left({x}-{4}\right)}{\left.{d}{x}\right.}=-{\frac{{{32}}}{{{3}}}}\)
0

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