Definite Integrals

$F(x)={\displaystyle \frac{x^2}{2}}$. ${\int }_0^{1}xdx=F(1)-F(0)={\displaystyle \frac{1}{2}}$.

$F(x)={\displaystyle \frac{x^3}{3}}$. $F(2)-F(1)={\displaystyle \frac{8}{3}}-{\displaystyle \frac{1}{3}}={\displaystyle \frac{7}{3}}$.

For an odd function $f$ and symmetric interval $[-a,a]$: ${\int }_{-a}^{a}f=0$.

${\int }_0^3{x}^{2}dx={\displaystyle \frac{27}{3}}=9$.

${\int }_0^1{e}^{x}dx=e^1-e^0=e-1$.

${\displaystyle \frac{1}{3}}{\int }_0^3{x}^{2}dx={\displaystyle \frac{1}{3}}\cdot 9=3$.

$u=x^2+1\Rightarrow du=2xdx$. When $x=0$, $u=1$; when $x=1$, $u=2$. ${\int }_1^2{u}^{3}du={\displaystyle \frac{16-1}{4}}={\displaystyle \frac{15}{4}}$.

$F(x)=\sin x$. $F(\pi \mathrm{/}2)-F(0)=1-0=1$.

$$\text{Area}={\int }_1^3(3x^2+2)dx$$

$$\int (3x^2+2)dx=x^3+2x$$

$$[x^3+2x{]}_1^3=(27+6)-(1+2)$$

$$=33-3=30$$

$\int \ln xdx=x\ln x-x+C$. Evaluate: $(e\cdot 1-e)-(1\cdot 0-1)=0-(-1)=1$.