Definite integral and the fundamental theorem

The indefinite integral builds a family of functions. The definite integral $\int_a^b f(x)\,dx$ builds a single number , the signed area between the graph of $f$ and the $x$-axis on the interval $[a, b]$. The Fundamental Theorem of Calculus says that, miraculously, these two objects are computed by the same operation.

Definition as a limit of a sum

Partition $[a, b]$ into $n$ equal subintervals of width $h = (b - a)/n$. Pick a sample point $\xi_i$ in the $i$-th subinterval. The Riemann sum is $S_n = \sum_{i=1}^n f(\xi_i) \cdot h.$ If the limit $\lim_{n \to \infty} S_n$ exists and is independent of the choice of sample points, we call it the definite integral: $\int_a^b f(x)\,dx = \lim_{n \to \infty} h \sum_{i=1}^{n} f(a + ih).$

Geometrically each rectangle has height $f(\xi_i)$ and width $h$, so the sum approximates the area under the curve. As $n \to \infty$, the approximation becomes exact.

The integral exists when $f$ is continuous on $[a, b]$, and more generally when $f$ has only finitely many jumps. Such $f$ are called Riemann integrable.

The Fundamental Theorem of Calculus

Part I (existence of antiderivatives). If $f$ is continuous on $[a, b]$, then the function $F(x) = \int_a^x f(t)\,dt$ is differentiable on $(a, b)$ with $F'(x) = f(x)$.

Part II (evaluation). If $F$ is any antiderivative of $f$ on $[a, b]$, then $\int_a^b f(x)\,dx = F(b) - F(a).$

This is the bridge. The hard limit-of-sums definition becomes the easy "compute antiderivative, plug in endpoints" procedure. The notation $\left[F(x)\right]_a^b = F(b) - F(a)$ is standard.

Why is the FTC true?

Sketch. Define $F(x) = \int_a^x f(t)\,dt$. Then $F(x + h) - F(x) = \int_x^{x+h} f(t)\,dt \approx h \cdot f(x)$ when $h$ is small (because $f$ is continuous, it is roughly constant near $x$). So $\frac{F(x+h) - F(x)}{h} \to f(x) \quad \text{as} \quad h \to 0.$ That is, $F'(x) = f(x)$. So $F$ is an antiderivative of $f$.

Now if $G$ is any other antiderivative, $G(x) = F(x) + C$. Then $G(b) - G(a) = F(b) - F(a) = \int_a^a f + \int_a^b f - \int_a^a f = \int_a^b f.$

So every antiderivative works for evaluation.

Definite integral as area

If $f \ge 0$ on $[a, b]$, then $\int_a^b f(x)\,dx$ is the area between the curve, the $x$-axis, and the vertical lines $x = a$ and $x = b$.

If $f$ changes sign, the integral gives signed area , regions below the $x$-axis contribute negatively. To find true geometric area, integrate $|f|$ instead, splitting at zeros of $f$.

Basic identities

Worked examples

Example 1. Evaluate $\int_0^1 x^2\,dx$ as a limit of a sum.

With $h = 1/n$ and $\xi_i = ih$: $S_n = h \sum_{i=1}^n (ih)^2 = h^3 \sum i^2 = \dfrac{1}{n^3} \cdot \dfrac{n(n+1)(2n+1)}{6}$. As $n \to \infty$ this $\to \dfrac{2}{6} = \dfrac{1}{3}$. Confirmed by FTC: $\left[\dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{3}$.

Example 2. $\int_0^{\pi/2} \cos x\,dx$.

Antiderivative is $\sin x$. So $\left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1$.

Example 3. $\int_1^e \dfrac{\ln x}{x}\,dx$.

Substitute $u = \ln x$, $du = dx/x$. When $x = 1$, $u = 0$; when $x = e$, $u = 1$. Integral $= \int_0^1 u\,du = 1/2$.

Example 4. $\int_0^1 \dfrac{dx}{1 + x^2}$.

$\left[\tan^{-1}x\right]_0^1 = \tan^{-1}1 - 0 = \pi/4$.

Example 5. Find $\dfrac{d}{dx}\int_0^{x^2} \cos(t^2)\,dt$.

Use the chain version of FTC Part I: $\dfrac{d}{dx} \int_{a}^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$. So the answer is $\cos(x^4) \cdot 2x = 2x\cos(x^4)$.

Example 6. Compute the area between $y = x^2$ and $y = x$ on $[0, 1]$.

The two curves meet at $x = 0$ and $x = 1$, with $x \ge x^2$ on $[0, 1]$. Area $= \int_0^1 (x - x^2)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$.

Try it yourself

1. $\int_0^2 (3x^2 + 1)\,dx$
2. $\int_{-1}^1 |x|\,dx$
3. $\int_0^{\pi} \sin x\,dx$
4. $\int_1^4 \sqrt{x}\,dx$
5. $\int_0^1 e^x\,dx$
6. $\int_1^2 \dfrac{1}{x^2}\,dx$
7. Use the limit-of-sums to evaluate $\int_0^2 x\,dx$.
8. $\int_0^{\pi/4} \sec^2 x\,dx$
9. $\int_0^{\pi/2} \sin^2 x\,dx$ (use double-angle)
10. Find $\dfrac{d}{dx}\int_x^{x^2} \sin t\,dt$.
11. $\int_0^1 \dfrac{x}{1+x^2}\,dx$
12. Find the area between $y = \sin x$ and the $x$-axis on $[0, 2\pi]$.
13. $\int_{-2}^{2} (x^3 + x)\,dx$ , predict the answer using symmetry.
14. $\int_0^a x(a - x)\,dx$ as a function of $a$.

Pitfalls and tricks

• No "$+C$" in definite integrals. The constant cancels in $F(b) - F(a)$.
• Change the limits under substitution. When you substitute $u = g(x)$, also change the limits to $u = g(a)$ and $u = g(b)$. (Or revert to $x$ first, then plug in original limits.)
• Signed vs. unsigned area. $\int_a^b f$ may be negative or zero even when $f$ is non-zero , it's only "area" when $f \ge 0$.
• Discontinuities matter. $\int_{-1}^{1} \dfrac{dx}{x^2}$ is not $\left[-\dfrac{1}{x}\right]_{-1}^1$ , the integrand blows up at $0$.