Section 9.3: Evaluation of Definite Integrals
The Fundamental Theorem of Calculus, recalled

If $F'(x) = f(x)$, then

$$\int_a^b f(x) dx = F(b) - F(a)$$

An example

Evaluate

$$\int_1^{10} \frac{(\ln (x))^4}{x} dx$$

A solution

Substitute $u = \ln(x)$ so that $du = \frac{dx}{x}$.

Thus,

$$\int \frac{(\ln(x))^4}{x} dx = \int u^4 du$$

$$= \frac{1}{5} u^5 + C$$

$$= \frac{1}{5} (\ln(x))^5 + C$$

Solution, continued

Thus,

$$\int_1^{10} \frac{(\ln (x))^4}{x} dx = \frac{1}{5} (\ln(10))^5 - \frac{1}{5} (\ln(1))^5$$

$$= \frac{1}{5} (\ln(10))^5$$

Substituting the limits as well

If we substitute $u = g(x)$ so that $\int f(x) dx = \int h(u) du$, then

$$\int_a^b f(x) dx = \int_{g(a)}^{g(b)} h(u) du$$

Indeed, if $H'(u) = h(u)$ and $f(x) = h(g(x)) g'(x)$, then $(H \circ g)'(x) = H'(g(x)) g'(x) = h(g(x)) g'(x) = f(x)$. Thus, by the fundamental theorem of calculus,

$$\int_{g(a)}^{g(b)} h(u) du = H(g(b)) - H(g(a))$$

$$= \int_a^b f(x) dx$$

Example

Evaluate

$$\int_1^2 x^2 e^{x^3} dx$$

Solution

Set $u = x^3$, then $du = 3 x^2 dx$. So,

$$\int_1^2 x^2 e^{x^3} dx = \frac{1}{3} \int_{u=1^3}^{u=2^3} e^u du$$

$$= \frac{1}{3} e^u \vert_{u=1}^{u=8}$$

$$= \frac{1}{3} e^8 - \frac{1}{3} e$$

Another example

Evaluate

$$\int_0^{\frac{\pi}{4}} \tan(x) dx$$

Solution

Write $\tan(x) = \frac{\sin(x)}{\cos(x)}$ Set $u = \cos(x)$. Then $du = - \sin(x) dx$.

Thus,

$$\int_0^{\frac{\pi}{4}} \tan(x) dx = -\int_{u=1}^{u=\frac{\sqrt{2}}{2}} \frac{du}{u}$$

$$= - \ln(u) \vert_{u=1}^{u=\frac{\sqrt{2}}{2}}$$

$$= - \ln(1) + \ln(\frac{\sqrt{2}}{2})$$

$$= \ln(2^{\frac{1}{2}}) - \ln(2)$$

$$= \frac{1}{2} \ln(2) - \ln(2)$$

$$= - \frac{1}{2} \ln(2)$$

Example

Compute

$$\int_0^\pi x \sin(x^2) + x^2 \cos(x) dx$$

Solution

Break the integral into two summands, $\int_0^\pi x \sin(x^2) dx$ and $\int_0^\pi x^2 \cos(x) dx$. We compute these separately. For the former, set $u = x^2$ so that $du = 2x dx$. Thus,

$$\int_{x=0}^{x = \pi} x \sin(x^2) dx = \frac{1}{2} \int_{u=0}^{u=\pi^2} \sin(u) du$$

$$= \frac{-1}{2} \cos(u) \vert_{u=0}^{u=\pi^2}$$

$$= \frac{1}{2} (1 - \cos(\pi^2))$$

Solution, continued

For the latter part, set $u = x^2$ and $dv = \cos(x) dx$. Then, $du = 2x dx$ and $v = \sin(x)$. So that

$$\int x^2 \cos(x) dx = x^2 \sin(x) - 2 \int x \sin(x) dx$$

Set $w = x$ and $dy = \sin(x) dx$. Then, $dw = dx$ and $y = -\cos(x)$. So,

$$\int x \sin(x) dx = -x \cos(x) + \int \cos(x) dx = -x \cos(x) + \sin(x) + C$$

Combining these,

$$\int x^2 \cos(x) dx = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C$$

Solution, continued

Evaluating,

$$\int_0^\pi x^2 \cos(x) dx = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) \vert_{x=0}^{x = \pi}$$

$$= \pi^2 \sin(\pi) + 2 \pi \cos(\pi)$$

$$- 2 \sin(\pi) - 0^2 \sin(0) - 2(0) \cos(0) + 2 \sin(0)$$

$$= 2 \pi$$

So,

$$\int_0^\pi (x \sin(x^2) + x^2 \cos(x)) dx = \frac{1}{2} (1 - \cos(\pi^2)) + 2 \pi$$