Definite integrals and areas practice questions — IB Maths AA

Let $u=\ln x$, so that $$du=\frac{1}{x}dx\mathrm{.}$$

The limits transform as $$x=1\Rightarrow u=0,x=e\Rightarrow u=1.$$

The integral becomes $${\int }_0^{1}udu={\left[\frac{u^2}{2}\right]}_0^1=\frac{1}{2}\mathrm{.}$$

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

The integral of a non-negative continuous function over $[a,b]$ is non-negative (representing the area under the curve).

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

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

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

$$=33-3=30$$

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

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

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

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

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

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