7.6 Definite Integrals

7.6 Definite Integrals

The definite integral of $f(x)$ from $a$ to $b$ is:

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

where $F$ is an antiderivative of $f$. This is the Fundamental Theorem of Calculus.

Properties

• $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

• $\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$

• $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$

• $\displaystyle\int_0^{2a} f(x)\,dx = \begin{cases} 2\displaystyle\int_0^a f(x)\,dx & \text{if } f(2a-x) = f(x) \\ 0 & \text{if } f(2a-x) = -f(x) \end{cases}$

Example

Evaluate $\displaystyle\int_0^{\pi/2} \sin x\,dx$.

$$\int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos\frac{\pi}{2} + \cos 0 = 0 + 1 = 1$$

Example (using property)

Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$.

Let $I = \displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$.

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$:

$$I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}\,dx$$

Adding both:

$$2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$$ $$\therefore \; I = \frac{\pi}{4}$$