Evaluation of Definite Integrals

In mathematics, the evaluation of definite integrals is a fundamental concept in calculus. It allows us to find the area under a curve, calculate accumulated change, and solve a wide range of problems in physics, engineering, and economics. Definite integrals are closely related to antiderivatives through the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) provides the primary method for evaluating definite integrals. It establishes a direct link between differentiation and integration. There are two parts to the theorem, but for evaluating definite integrals, we primarily use the second part.

FTC, Part 2 (The Evaluation Theorem)

If a function f is continuous on the closed interval [a, b] and F is any antiderivative of f on [a, b] (meaning F'(x) = f(x)), then the definite integral of f from a to b is given by:

$$ \int_{a}^{b} f(x) \, dx = F(b) - F(a) $$

This is often written using the notation [F(x)]_a^b or F(x) |_a^b.

Steps for Evaluating Definite Integrals using FTC:

1. Find an antiderivative, F(x), of the integrand, f(x). This means finding a function whose derivative is f(x). Remember that the constant of integration, C, is not needed when evaluating definite integrals because it cancels out: (F(b) + C) - (F(a) + C) = F(b) - F(a).
2. Evaluate the antiderivative at the upper limit of integration, b, to get F(b).
3. Evaluate the antiderivative at the lower limit of integration, a, to get F(a).
4. Subtract the value at the lower limit from the value at the upper limit: F(b) - F(a).

Examples of Basic Definite Integrals

Example 1: Polynomial Function

Evaluate $$ \int_{1}^{3} x^2 \, dx $$

1. Find the antiderivative of f(x) = x². Using the power rule for integration, the antiderivative is $$ F(x) = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} $$.
2. Evaluate F(x) at the upper limit, b = 3: $$ F(3) = \frac{3^3}{3} = \frac{27}{3} = 9 $$.
3. Evaluate F(x) at the lower limit, a = 1: $$ F(1) = \frac{1^3}{3} = \frac{1}{3} $$.
4. Subtract: $$ F(3) - F(1) = 9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} $$.

So, $$ \int_{1}^{3} x^2 \, dx = \frac{26}{3} $$.

Example 2: Trigonometric Function

Evaluate $$ \int_{0}^{\pi/2} \cos(x) \, dx $$

1. The antiderivative of f(x) = cos(x) is $$ F(x) = \sin(x) $$.
2. Evaluate at the upper limit, b = π/2: $$ F(\pi/2) = \sin(\pi/2) = 1 $$.
3. Evaluate at the lower limit, a = 0: $$ F(0) = \sin(0) = 0 $$.
4. Subtract: $$ F(\pi/2) - F(0) = 1 - 0 = 1 $$.

Therefore, $$ \int_{0}^{\pi/2} \cos(x) \, dx = 1 $$.

Integration Techniques for Definite Integrals

When the integrand is more complex, we often need to use various integration techniques before applying the Fundamental Theorem of Calculus. These techniques transform the integrand into a form that is easier to integrate.

1. Substitution Rule (u-Substitution)

The substitution rule for definite integrals is a variation of the rule for indefinite integrals. If u = g(x), then du = g'(x) dx. When we change the variable from x to u, we must also change the limits of integration.

If u = g(x), then the new lower limit is g(a) and the new upper limit is g(b). The integral becomes:

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

Alternatively, you can find the antiderivative in terms of x and then substitute the original limits.

Example 3: Using Substitution

Evaluate $$ \int_{0}^{1} x \sqrt{1+x^2} \, dx $$

Let $$ u = 1+x^2 $$. Then $$ du = 2x \, dx $$, which means $$ x \, dx = \frac{1}{2} \, du $$.

Change the limits of integration:

• When $$ x = 0 $$, $$ u = 1 + 0^2 = 1 $$.
• When $$ x = 1 $$, $$ u = 1 + 1^2 = 2 $$.

The integral becomes: $$ \int_{1}^{2} \sqrt{u} \left(\frac{1}{2} \, du\right) = \frac{1}{2} \int_{1}^{2} u^{1/2} \, du $$

Now, evaluate using FTC: $$ \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{1}^{2} = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} $$ $$ = \frac{1}{3} \left[ u^{3/2} \right]_{1}^{2} = \frac{1}{3} (2^{3/2} - 1^{3/2}) $$ $$ = \frac{1}{3} (2\sqrt{2} - 1) $$

So, $$ \int_{0}^{1} x \sqrt{1+x^2} \, dx = \frac{2\sqrt{2} - 1}{3} $$.

2. Integration by Parts

The formula for integration by parts is derived from the product rule for differentiation. For indefinite integrals, it is $$ \int u \, dv = uv - \int v \, du $$. For definite integrals, the formula becomes:

$$ \int_{a}^{b} u \, dv = [uv]_{a}^{b} - \int_{a}^{b} v \, du $$

This means $$ \int_{a}^{b} u(x) v'(x) \, dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} v(x) u'(x) \, dx $$.

The choice of u and dv is crucial. A common mnemonic is LIATE (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential) to decide which function to choose as u, as it is generally easier to differentiate u and integrate dv.

Example 4: Using Integration by Parts

Evaluate $$ \int_{0}^{1} x e^x \, dx $$

Let $$ u = x $$ (Algebraic) and $$ dv = e^x \, dx $$ (Exponential). Then $$ du = dx $$ and $$ v = \int e^x \, dx = e^x $$.

Apply the integration by parts formula for definite integrals: $$ \int_{0}^{1} x e^x \, dx = [x e^x]_{0}^{1} - \int_{0}^{1} e^x \, dx $$

Evaluate the first term: $$ [x e^x]_{0}^{1} = (1 \cdot e^1) - (0 \cdot e^0) = e - 0 = e $$

Evaluate the second term: $$ \int_{0}^{1} e^x \, dx = [e^x]_{0}^{1} = e^1 - e^0 = e - 1 $$

Combine the results: $$ \int_{0}^{1} x e^x \, dx = e - (e - 1) = e - e + 1 = 1 $$

Thus, $$ \int_{0}^{1} x e^x \, dx = 1 $$.

3. Trigonometric Substitution

This technique is used for integrals involving expressions of the form $$ \sqrt{a^2 \pm x^2} $$ or $$ \sqrt{x^2 - a^2} $$. We substitute x with a trigonometric function (like a sin(θ), a tan(θ), or a sec(θ)) to simplify the expression under the square root using trigonometric identities.

• For $$ \sqrt{a^2 - x^2} $$, let $$ x = a \sin(\theta) $$. Then $$ \sqrt{a^2 - x^2} = a \cos(\theta) $$.
• For $$ \sqrt{a^2 + x^2} $$, let $$ x = a \tan(\theta) $$. Then $$ \sqrt{a^2 + x^2} = a \sec(\theta) $$.
• For $$ \sqrt{x^2 - a^2} $$, let $$ x = a \sec(\theta) $$. Then $$ \sqrt{x^2 - a^2} = a \tan(\theta) $$.

When using trigonometric substitution for definite integrals, remember to change the limits of integration from x-values to θ-values.

Example 5: Trigonometric Substitution

Evaluate $$ \int_{0}^{1} \frac{1}{\sqrt{4-x^2}} \, dx $$

Here, $$ a^2 = 4 $$, so $$ a = 2 $$. The form is $$ \sqrt{a^2 - x^2} $$. Let $$ x = 2 \sin(\theta) $$. Then $$ dx = 2 \cos(\theta) \, d\theta $$. Also, $$ \sqrt{4-x^2} = \sqrt{4 - 4 \sin^2(\theta)} = \sqrt{4 \cos^2(\theta)} = 2 \cos(\theta) $$.

Change the limits of integration:

• When $$ x = 0 $$, $$ 0 = 2 \sin(\theta) \implies \sin(\theta) = 0 \implies \theta = 0 $$.
• When $$ x = 1 $$, $$ 1 = 2 \sin(\theta) \implies \sin(\theta) = 1/2 \implies \theta = \pi/6 $$.

The integral becomes: $$ \int_{0}^{\pi/6} \frac{1}{2 \cos(\theta)} (2 \cos(\theta) \, d\theta) = \int_{0}^{\pi/6} 1 \, d\theta $$

Evaluate: $$ [\theta]_{0}^{\pi/6} = \frac{\pi}{6} - 0 = \frac{\pi}{6} $$

So, $$ \int_{0}^{1} \frac{1}{\sqrt{4-x^2}} \, dx = \frac{\pi}{6} $$.

4. Partial Fractions

This method is used to integrate rational functions (a ratio of polynomials), P(x)/Q(x), where the denominator Q(x) can be factored. The rational function is decomposed into a sum of simpler fractions whose integrals are easier to find.

Example 6: Partial Fractions

Evaluate $$ \int_{2}^{3} \frac{1}{x^2 - 1} \, dx $$

First, factor the denominator: $$ x^2 - 1 = (x-1)(x+1) $$. Decompose the integrand using partial fractions: $$ \frac{1}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} $$ Multiply by $$ (x-1)(x+1) $$: $$ 1 = A(x+1) + B(x-1) $$ To find A, set $$ x = 1 $$: $$ 1 = A(1+1) + B(1-1) \implies 1 = 2A \implies A = 1/2 $$. To find B, set $$ x = -1 $$: $$ 1 = A(-1+1) + B(-1-1) \implies 1 = -2B \implies B = -1/2 $$.

So, $$ \frac{1}{x^2 - 1} = \frac{1/2}{x-1} - \frac{1/2}{x+1} $$.

Now, integrate: $$ \int_{2}^{3} \left( \frac{1/2}{x-1} - \frac{1/2}{x+1} \right) \, dx = \frac{1}{2} \int_{2}^{3} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) \, dx $$ $$ = \frac{1}{2} [\ln|x-1| - \ln|x+1|]_{2}^{3} $$ $$ = \frac{1}{2} \left[ \ln\left|\frac{x-1}{x+1}\right| \right]_{2}^{3} $$

Evaluate at the limits: $$ \frac{1}{2} \left( \ln\left|\frac{3-1}{3+1}\right| - \ln\left|\frac{2-1}{2+1}\right| \right) $$ $$ = \frac{1}{2} \left( \ln\left|\frac{2}{4}\right| - \ln\left|\frac{1}{3}\right| \right) $$ $$ = \frac{1}{2} \left( \ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right) \right) $$ $$ = \frac{1}{2} \left( -\ln(2) - (-\ln(3)) \right) $$ $$ = \frac{1}{2} (\ln(3) - \ln(2)) = \frac{1}{2} \ln\left(\frac{3}{2}\right) $$

Therefore, $$ \int_{2}^{3} \frac{1}{x^2 - 1} \, dx = \frac{1}{2} \ln\left(\frac{3}{2}\right) $$.

Improper Integrals

Improper integrals are definite integrals where either the interval of integration is infinite, or the integrand has an infinite discontinuity within the interval of integration. They are evaluated using limits.

Type 1: Infinite Interval of Integration

If f is continuous on $$ [a, \infty) $$, then $$ \int_{a}^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_{a}^{t} f(x) \, dx $$. If f is continuous on $$ (-\infty, b] $$, then $$ \int_{-\infty}^{b} f(x) \, dx = \lim_{t \to -\infty} \int_{t}^{b} f(x) \, dx $$. If both $$ \int_{a}^{\infty} f(x) \, dx $$ and $$ \int_{-\infty}^{a} f(x) \, dx $$ converge for some number a, then $$ \int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{a} f(x) \, dx + \int_{a}^{\infty} f(x) \, dx $$.

Example 7: Infinite Interval

Evaluate $$ \int_{1}^{\infty} \frac{1}{x^2} \, dx $$

1. Rewrite as a limit: $$ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_{1}^{t} x^{-2} \, dx $$
2. Find the antiderivative: $$ \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} $$.
3. Evaluate the definite integral: $$ \int_{1}^{t} x^{-2} \, dx = \left[-\frac{1}{x}\right]_{1}^{t} = -\frac{1}{t} - \left(-\frac{1}{1}\right) = 1 - \frac{1}{t} $$.
4. Take the limit: $$ \lim_{t \to \infty} \left(1 - \frac{1}{t}\right) = 1 - 0 = 1 $$.

Thus, $$ \int_{1}^{\infty} \frac{1}{x^2} \, dx = 1 $$. The integral converges.

Type 2: Infinite Discontinuity

If f is continuous on $$ [a, b) $$ and has an infinite discontinuity at b, then $$ \int_{a}^{b} f(x) \, dx = \lim_{t \to b^-} \int_{a}^{t} f(x) \, dx $$. If f is continuous on $$ (a, b] $$ and has an infinite discontinuity at a, then $$ \int_{a}^{b} f(x) \, dx = \lim_{t \to a^+} \int_{t}^{b} f(x) \, dx $$. If f has a discontinuity at c in $$ (a, b) $$, then $$ \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx $$, provided both integrals on the right converge.

Example 8: Infinite Discontinuity

Evaluate $$ \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx $$

1. The integrand $$ \frac{1}{\sqrt{x}} $$ has an infinite discontinuity at $$ x = 0 $$. Rewrite as a limit: $$ \int_{0}^{1} x^{-1/2} \, dx = \lim_{t \to 0^+} \int_{t}^{1} x^{-1/2} \, dx $$
2. Find the antiderivative: $$ \int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x} $$.
3. Evaluate the definite integral: $$ \int_{t}^{1} x^{-1/2} \, dx = [2\sqrt{x}]_{t}^{1} = 2\sqrt{1} - 2\sqrt{t} = 2 - 2\sqrt{t} $$.
4. Take the limit: $$ \lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2 - 2\sqrt{0} = 2 $$.

Therefore, $$ \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx = 2 $$. The integral converges.

Numerical Integration

Sometimes, it is impossible or impractical to find an exact antiderivative for an integrand. In such cases, we resort to numerical methods to approximate the value of the definite integral.

1. Trapezoidal Rule

This method approximates the area under the curve by dividing the interval $$ [a, b] $$ into n subintervals of equal width $$ \Delta x = \frac{b-a}{n} $$. The area is approximated by summing the areas of trapezoids formed in each subinterval.

The formula is:

$$ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)] $$

where $$ x_i = a + i \Delta x $$.

2. Simpson's Rule

Simpson's Rule provides a more accurate approximation by using parabolic segments instead of straight lines to approximate the curve. It requires an even number of subintervals, n.

The formula is:

$$ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)] $$

where $$ \Delta x = \frac{b-a}{n} $$ and n is an even integer.

Applications of Definite Integrals

Definite integrals have numerous applications beyond finding areas:

• Calculating Volumes: Volumes of solids of revolution (disk method, washer method, shell method) and other solids.
• Finding Arc Length: The length of a curve over a given interval.
• Calculating Surface Area: The area of a surface of revolution.
• Work Done: In physics, the work done by a variable force over a distance.
• Center of Mass and Moments: Determining the physical properties of objects.
• Probability: Calculating probabilities for continuous random variables.
• Fluid Pressure and Force: Calculating the force exerted by fluids on submerged surfaces.