Use the form of the definition of the integral given in Theorem 4 to evaluate...

Step 1
To evaluate the integral: ${\displaystyle {\int }_0^2(2x-x^3)dx}$
Evaluating the integral.
${\displaystyle {\int }_0^2(2x-x^3)dx={\int }_0^{2}2xdx-{\int }_0^2{x}^{3}dx}$
${\displaystyle ={(2\cdot \frac{x^2}{2})}_0^2-{\left(\frac{x^4}{4}\right)}_0^2}$
${\displaystyle ={\left(x^2\right)}_0^2-(\frac{2^2}{4}-0)}$
=(4-0)-(1-0)
=4-1
=3
Step 2
Hence, required answer is 3.

${\displaystyle \int (2x-x^3)dx}$
Lets

We represent the original integral as a sum of tabular integrals:
$\int (2\ast x-x^3)\ast dx=\int -x^3\ast dx+\int 2\ast x\ast dx$
$\int (-x^3)\ast dx$
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:
$\int 2\ast x\ast dx=x^2+C$
$\int (2\ast x-x^3)\ast dx=-\frac{x^4}{4}+x^2+C$
Let's calculate the definite integral:
${\int }_0^2(2\ast x-x^3)\ast dx=(-\frac{x^4}{4}+x^2{)}_0^2$
F(2)=0
F(0)=0
I=0-(0)=0