Definite integrals and areas practice questions — IB Maths AA

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

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

The integral becomes $$\int_0^1 u\,du=\left[\frac{u^2}{2}\right]_0^1=\frac{1}{2}.$$

$\int \ln x \, dx = x \ln x - x + C$. 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).

$$\text{Area} = \int_1^3 \left(3x^2 + 2\right)\,dx$$

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

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

$$= 33 - 3 = 30$$

$\dfrac{1}{3}\int_0^3 x^2 \, dx = \dfrac{1}{3} \cdot 9 = 3$.

$\int_0^3 x^2 \, dx = \dfrac{27}{3} = 9$.

$\int_0^1 e^x \, dx = e^1 - e^0 = e - 1$.

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

$u = x^2 + 1 \Rightarrow du = 2x \, dx$. When $x = 0$, $u = 1$; when $x = 1$, $u = 2$. $\int_1^2 u^3 \, du = \dfrac{16 - 1}{4} = \dfrac{15}{4}$.

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