 David Troyer

2021-12-16

Use the form of the definition of the integral given in
Theorem 4 to evaluate the integral.
${\int }_{0}^{2}\left(2x-{x}^{3}\right)dx$ temzej9

Expert

Step 1
To evaluate the integral: ${\int }_{0}^{2}\left(2x-{x}^{3}\right)dx$
Evaluating the integral.
${\int }_{0}^{2}\left(2x-{x}^{3}\right)dx={\int }_{0}^{2}2xdx-{\int }_{0}^{2}{x}^{3}dx$
$={\left(2\cdot \frac{{x}^{2}}{2}\right)}_{0}^{2}-{\left(\frac{{x}^{4}}{4}\right)}_{0}^{2}$
$={\left({x}^{2}\right)}_{0}^{2}-\left(\frac{{2}^{2}}{4}-0\right)$
=(4-0)-(1-0)
=4-1
=3
Step 2 levurdondishav4

Expert

$\int \left(2x-{x}^{3}\right)dx$
Lets nick1337

Expert

We represent the original integral as a sum of tabular integrals:
$\int \left(2\ast x-{x}^{3}\right)\ast dx=\int -{x}^{3}\ast dx+\int 2\ast x\ast dx$
$\int \left(-{x}^{3}\right)\ast dx$
This is a tabular integral:
$\int -{x}^{3}\ast dx=-\frac{{x}^{4}}{4}+C$
$\int 2\ast x\ast dx$
This is a tabular integral:
$\int 2\ast x\ast dx={x}^{2}+C$
$\int \left(2\ast x-{x}^{3}\right)\ast dx=-\frac{{x}^{4}}{4}+{x}^{2}+C$
Let's calculate the definite integral:
${\int }_{0}^{2}\left(2\ast x-{x}^{3}\right)\ast dx=\left(-\frac{{x}^{4}}{4}+{x}^{2}{\right)}_{0}^{2}$
F(2)=0
F(0)=0
I=0-(0)=0

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