Evaluating Definite Integrals

A definite integral calculates the exact accumulation of a quantity, such as the area under a curve between two points. To evaluate a definite integral, we use antiderivatives and the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration. It states that if $f(x)$ is a continuous function on the interval $[a, b]$ and $F(x)$ is any antiderivative of $f(x)$ (meaning $F'(x) = f(x)$), then the definite integral is evaluated as:

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

This is often written with evaluation brackets: $\left[ F(x) \right]_a^b$.

Key Properties of Definite Integrals

Understanding these properties makes evaluating complex integrals much simpler:

• Linearity: You can split addition/subtraction and pull out constants. $\int_a^b [c f(x) \pm d g(x)] \, dx = c \int_a^b f(x) \, dx \pm d \int_a^b g(x) \, dx$
• Interval Splitting: You can break an integral into adjacent intervals. $\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$
• Reversing Limits: Swapping the upper and lower bounds changes the sign of the integral. $\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx$
• Zero Length Interval: Integrating from a point to the exact same point yields zero. $\int_a^a f(x) \, dx = 0$

Example Problems

Let's apply the Fundamental Theorem and the properties to evaluate some integrals.

Example 1: Polynomials and Roots

Evaluate $\int_1^4 (2x - \sqrt{x}) \, dx$

1. First, rewrite the square root as a fractional exponent: $\int_1^4 (2x - x^{1/2}) \, dx$
2. Find the antiderivative $F(x)$ using the power rule ($\int x^n \, dx = \frac{x^{n+1}}{n+1}$): $F(x) = x^2 - \frac{x^{3/2}}{3/2} = x^2 - \frac{2}{3}x^{3/2}$
3. Evaluate $F(b) - F(a)$ from $x=1$ to $x=4$: $\left[ x^2 - \frac{2}{3}x^{3/2} \right]_1^4$
4. Plug in the upper limit ($4$) and lower limit ($1$): $\left( 4^2 - \frac{2}{3}(4)^{3/2} \right) - \left( 1^2 - \frac{2}{3}(1)^{3/2} \right)$ $\left( 16 - \frac{2}{3}(8) \right) - \left( 1 - \frac{2}{3} \right)$ $\left( 16 - \frac{16}{3} \right) - \left( \frac{1}{3} \right)$ $\frac{48}{3} - \frac{16}{3} - \frac{1}{3} = \frac{31}{3}$

Example 2: Trigonometric Functions

Evaluate $\int_0^{\pi/2} (3\cos x + 2\sin x) \, dx$

1. Find the antiderivative. Remember that $\int \cos x \, dx = \sin x$ and $\int \sin x \, dx = -\cos x$: $F(x) = 3\sin x - 2\cos x$
2. Evaluate $F(b) - F(a)$ from $0$ to $\pi/2$: $\left[ 3\sin x - 2\cos x \right]_0^{\pi/2}$
3. Plug in the upper limit ($\pi/2$) and lower limit ($0$): $\left( 3\sin\left(\frac{\pi}{2}\right) - 2\cos\left(\frac{\pi}{2}\right) \right) - \left( 3\sin(0) - 2\cos(0) \right)$
4. Simplify using exact trigonometric values ($\sin(\pi/2)=1$, $\cos(\pi/2)=0$, $\sin(0)=0$, $\cos(0)=1$): $\left( 3(1) - 2(0) \right) - \left( 3(0) - 2(1) \right)$ $(3 - 0) - (0 - 2) = 3 + 2 = 5$