Definite Integral and Fundamental Theorem of Calculus

Introduction to Definite Integrals

While the indefinite integral $\int f(x)\,dx$ represents a family of anti-derivatives differing by an arbitrary constant $C$, a definite integral gives a single numerical value.

The definite integral of a function $f(x)$ from $x=a$ to $x=b$ is written as: $$\int_a^b f(x)\,dx$$ Here:

• $a$ is the lower limit
• $b$ is the upper limit

If $f(x) \ge 0$ for all $x \in [a,b]$, then geometrically $\int_a^b f(x)\,dx$ represents the area bounded by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$.

If $f(x)$ takes negative values on part of the interval, then the definite integral represents signed area: area above the x-axis is taken positive and area below the x-axis is taken negative.

Definite Integral as the Limit of a Sum

The definite integral arises from the idea of approximating area under a curve by dividing the interval into many thin rectangles and then adding their areas.

Let $f(x)$ be continuous on $[a,b]$. Divide the interval into $n$ equal parts, each of width $$h = \frac{b-a}{n}.$$ Then the points of division are: $$a,\ a+h,\ a+2h,\ \dots,\ a+nh=b.$$

Using left endpoints, the definite integral is defined by the limit: $$\int_a^b f(x)\,dx = \lim_{n\to\infty} h\,[f(a)+f(a+h)+f(a+2h)+\dots+f(a+(n-1)h)].$$

Equivalently, in sigma notation: $$\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{r=0}^{n-1} f(a+rh)\,h, \qquad h=\frac{b-a}{n}.$$

This is called a Riemann sum. In practice, however, evaluating definite integrals directly from this definition is cumbersome, so we use the Fundamental Theorem of Calculus.

First Fundamental Theorem of Calculus

Let $f$ be continuous on the closed interval $[a,b]$. Define the area function $$A(x)=\int_a^x f(t)\,dt.$$ Here $t$ is called a dummy variable; it is used only inside the integral.

First Fundamental Theorem of Calculus: If $A(x)=\int_a^x f(t)\,dt$, then $$A'(x)=f(x).$$

This theorem shows that differentiation and integration are inverse processes. In other words, if you first accumulate area under the curve and then differentiate that area function, you recover the original function.

Second Fundamental Theorem of Calculus

This theorem is the practical tool used to evaluate definite integrals.

Let $f$ be continuous on $[a,b]$ and let $F$ be any anti-derivative of $f$, that is, $$F'(x)=f(x).$$ Then: $$\int_a^b f(x)\,dx = [F(x)]_a^b = F(b)-F(a).$$

Working rule

1. Find an anti-derivative $F(x)$ of $f(x)$.
2. Evaluate $F(b)$ at the upper limit.
3. Evaluate $F(a)$ at the lower limit.
4. Subtract: $F(b)-F(a)$.

There is no need to write the constant of integration $C$, because it cancels automatically: $$[F(b)+C]-[F(a)+C]=F(b)-F(a).$$

Evaluation of Definite Integrals by Substitution

When using substitution in a definite integral, it is essential to change the limits according to the new variable.

Suppose we want to evaluate: $$\int_a^b f(g(x))g'(x)\,dx$$ Make the substitution: $$t=g(x), \qquad dt=g'(x)\,dx.$$ Then the limits change as follows:

• when $x=a$, the new lower limit becomes $t=g(a)$
• when $x=b$, the new upper limit becomes $t=g(b)$

So the integral becomes: $$\int_{g(a)}^{g(b)} f(t)\,dt.$$

A major advantage of this method is that once the limits are changed properly, you do not need to substitute back to the variable $x$ at the end.

Example 1: Evaluating a Polynomial Definite Integral

Evaluate: $\int_1^3 (x^2+2x)\,dx$

Solution: Step 1: Find an anti-derivative of $x^2+2x$. $$\int (x^2+2x)\,dx = \frac{x^3}{3}+x^2$$ So we take: $$F(x)=\frac{x^3}{3}+x^2$$

Step 2: Apply the Second Fundamental Theorem of Calculus. $$\int_1^3 (x^2+2x)\,dx = \left[\frac{x^3}{3}+x^2\right]_1^3$$

Step 3: Evaluate at the upper limit $x=3$. $$F(3)=\frac{27}{3}+9=9+9=18$$

Step 4: Evaluate at the lower limit $x=1$. $$F(1)=\frac{1}{3}+1=\frac{4}{3}$$

Step 5: Subtract. $$F(3)-F(1)=18-\frac{4}{3}=\frac{54-4}{3}=\frac{50}{3}$$

Answer: $\frac{50}{3}$

Example 2: Evaluating a Trigonometric Definite Integral

Evaluate: $\int_0^{\pi/4} \sec^2 x\,dx$

Solution: Step 1: Recall the anti-derivative: $$\int \sec^2 x\,dx = \tan x$$ So we take: $$F(x)=\tan x$$

Step 2: Apply the limits. $$\int_0^{\pi/4} \sec^2 x\,dx = [\tan x]_0^{\pi/4}$$

Step 3: Evaluate. $$= \tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1$$

Answer: $1$

Example 3: Definite Integral of an Exponential Function

Evaluate: $\int_0^1 e^x\,dx$

Solution: Step 1: The anti-derivative of $e^x$ is itself. $$F(x)=e^x$$

Step 2: Apply the limits. $$\int_0^1 e^x\,dx = [e^x]_0^1$$

Step 3: Evaluate at the endpoints. $$= e^1-e^0=e-1$$

Answer: $e-1$

Example 4: Definite Integral by Substitution

Evaluate: $\int_0^1 \frac{2x}{1+x^2}\,dx$

Solution: Step 1: Use substitution. Let: $$t=1+x^2$$ Then: $$dt=2x\,dx$$

Step 2: Change the limits. When $x=0$, $$t=1+0^2=1$$ When $x=1$, $$t=1+1^2=2$$

Step 3: Rewrite the integral completely in terms of $t$. $$\int_1^2 \frac{1}{t}\,dt$$

Step 4: Integrate. $$\int \frac{1}{t}\,dt = \ln|t|$$ So, $$\int_1^2 \frac{1}{t}\,dt = [\ln|t|]_1^2$$

Step 5: Evaluate the limits. $$= \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$$

Answer: $\ln 2$

Example 5: Substitution with Trigonometric Functions

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

Solution: Step 1: Use substitution. Let: $$t=\sin x$$ Then: $$dt=\cos x\,dx$$

Step 2: Change the limits. When $x=0$, $$t=\sin 0=0$$ When $x=\pi/2$, $$t=\sin\left(\frac{\pi}{2}\right)=1$$

Step 3: Rewrite the integral. $$\int_0^1 t^3\,dt$$

Step 4: Integrate. $$\int t^3\,dt = \frac{t^4}{4}$$ So, $$\left[\frac{t^4}{4}\right]_0^1 = \frac{1^4}{4}-\frac{0^4}{4}=\frac{1}{4}$$

Answer: $\frac{1}{4}$