If differentiation is about breaking things apart (finding the rate of change), integration is about putting small pieces together to find the total. In physics: Work done by a variable force is $\int F\, dx$. Finding electric field lines, potential energy, center of mass — all these require integration. It is the other half of the calculus engine.
Integration is the inverse operation of differentiation. If $\frac{d}{dx}[F(x)] = f(x)$, then:
where $C$ is the constant of integration.
Since $\frac{d}{dx}(x^2) = 2x$ AND $\frac{d}{dx}(x^2 + 5) = 2x$... we write $\int 2x\, dx = x^2 + C$ to represent any possible constant.
Area under $y = x^2$ from $x = 0$ to $x = 3$ (shaded region)
Geometrically, the definite integral $\int_a^b f(x)\, dx$ gives the area enclosed between the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$.
| Function $f(x)$ | Integral $\int f(x)\, dx$ |
|---|---|
| $x^n$ ($n \neq -1$) | $\frac{x^{n+1}}{n+1} + C$ |
| $1/x$ | $\ln|x| + C$ |
| $e^x$ | $e^x + C$ |
| $\sin x$ | $-\cos x + C$ |
| $\cos x$ | $\sin x + C$ |
Example: Integrate $\int (4x^3 - 2x + 5)\, dx$
$= 4 \cdot \frac{x^4}{4} - 2 \cdot \frac{x^2}{2} + 5x = \mathbf{x^4 - x^2 + 5x + C}$
If the integral has a composite function, use $u$-substitution.
Example: Integrate $\int 2x(x^2 + 1)^5\, dx$
Let $u = x^2 + 1 \implies du = 2x\, dx$. Integral becomes $\int u^5 du = \mathbf{\frac{(x^2+1)^6}{6} + C}$.
Example (Work Done): $F = 2x + 3$. Find work from $x=0$ to $x=4$.
$W = \int_0^4 (2x+3) dx = [x^2 + 3x]_0^4 = 16 + 12 = \mathbf{28\text{ J}}$
| Q | Answer & Method |
|---|---|
| 1 | $x^5 - x^3 + 7x + C$ |
| 2 | $\frac{2}{3}x^{3/2} - 1/x + C$ |
| 3 | $-3\cos x - 2\sin x + C$ |
| 4 | $[x^3-x^2+x]_1^4 = 52-1 = \mathbf{51}$ |
| 5 | $[\sin x]_0^{\pi/2} = \mathbf{1}$ |
| 6 | $[x^3/3]_0^3 = \mathbf{9}$ |
| 7 | $[2x^2-x]_1^3 = 15-1 = \mathbf{14\text{ J}}$ |
| 8 | $[t^3+t^2]_0^3 = 27+9 = \mathbf{36\text{ m}}$ |
| 9 | $[t^2+3t]_0^4 = 16+12 = \mathbf{28\text{ C}}$ |