Question

If $f\left(x\right)=\{\begin{array}{c}2x^2+1,x\le 1\\ 4x^2-1,x>1\end{array}$, then ${\int }_0^{2}f\left(x\right)dx$ is

• A. $10$
• B. $50/3$
• C. $1/3$
• D. $47/2$

Answer 1

The correct option is A $10$

$f\left(x\right)=\{\begin{array}{c}2x^2+1,x\le 1\\ 4x^2-1,x>1\end{array}$

from 0 to 1 we have $2x^2+1$ and from 1 to 2 we have $4x^2-1$

$={\int }_0^1(2x^2+1)dx+{\int }_1^2(4x^2-1)dx$

$={[\frac{2x^3}{3}+x]}_0^1+{[\frac{4x^3}{3}-x]}_1^2$

$=\frac{2}{3}+1-0-0+\frac{4{\left(2\right)}^3}{3}-2-(\frac{4}{3}-1)$

$=\frac{2}{3}+1+\frac{32}{3}-2-\frac{4}{3}+1$

$=\frac{2+32-4}{3}$

$=\frac{30}{3}$

=10.