# Evaluate the definite integral using integration by parts. \int_{0}^{2}z(z-2)^{4}dz

Evaluate the definite integral using integration by parts.
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{z}{\left({z}-{2}\right)}^{{{4}}}{\left.{d}{z}\right.}$$

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Step 1
We need to Evaluate the definite integral using integration by parts of the following integral.
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{z}{\left({z}-{2}\right)}^{{{4}}}{\left.{d}{z}\right.}$$
Note:
Integration by parts:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{v}{\frac{{{d}{u}}}{{{\left.{d}{z}\right.}}}}{\left.{d}{z}\right.}={{\left[{u}{v}-\int{u}{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}{\left.{d}{z}\right.}\right]}_{{{a}}}^{{{b}}}}$$
Step 2
So,
For the given integral,
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{z}{\left({z}-{2}\right)}^{{{4}}}{\left.{d}{z}\right.}={{\left[{z}{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}-\int{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}\times{\frac{{{d}}}{{{\left.{d}{z}\right.}}}}{\left({z}\right)}{\left.{d}{z}\right.}\right]}_{{{0}}}^{{{2}}}}$$
$$\displaystyle={{\left[{z}{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}-\int{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}{\left.{d}{z}\right.}\right]}_{{{0}}}^{{{2}}}}$$
$$\displaystyle={{\left[{z}{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}-{\frac{{{\left({z}-{2}\right)}^{{{6}}}}}{{{5}\times{6}}}}\right]}_{{{0}}}^{{{2}}}}$$
$$\displaystyle={0}-{\left({0}-{\frac{{{\left({0}-{2}\right)}^{{{6}}}}}{{{5}\times{6}}}}\right)}$$
$$\displaystyle={\frac{{{32}}}{{{15}}}}$$
In the exact form: $$\displaystyle{\int_{{{0}}}^{{{2}}}}{z}{\left({z}-{2}\right)}^{{{4}}}{\left.{d}{z}\right.}={{\left[{z}{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}-\int{\frac{{{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}\times{\frac{{{d}}}{{{\left.{d}{z}\right.}}}}{\left({z}\right)}{\left.{d}{z}\right.}\right]}_{{{0}}}^{{{2}}}}$$
And the value of the integral is $$\displaystyle{\int_{{{0}}}^{{{2}}}}{z}{\left({z}-{2}\right)}^{{{4}}}{\left.{d}{z}\right.}={\frac{{{32}}}{{{15}}}}$$.
###### Have a similar question?
Marian Tucker
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({z}\right)}}}={z}{\left({z}-{2}\right)}^{{{4}}}$$. Find its integral.
$$\displaystyle{\frac{{{\left({z}-{2}\right)}^{{{6}}}}}{{{6}}}}+{\frac{{{2}{\left({z}-{2}\right)}^{{{5}}}}}{{{5}}}}{{\mid}_{{{0}}}^{{{2}}}}$$
Step 3: Since $$\displaystyle{F}{\left({z}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(2)−F(0):
$$\displaystyle{\left({\frac{{{\left({2}-{2}\right)}^{{{6}}}}}{{{6}}}}+{\frac{{{2}{\left({2}-{2}\right)}^{{{5}}}}}{{{5}}}}\right)}-{\left({\frac{{{\left({0}-{2}\right)}^{{{6}}}}}{{{6}}}}+{\frac{{{2}{\left({0}-{2}\right)}^{{{5}}}}}{{{5}}}}\right)}$$
Step 4: Simplify.
$$\displaystyle{\frac{{{32}}}{{{15}}}}$$