Integration I

Integration undoes differentiation. Everything in this topic follows from reading the differentiation rules backwards — and from remembering the constant that differentiation destroyed.

Antidifferentiation

If $\dfrac{dF}{dx} = f(x)$, then $F$ is an antiderivative of $f$, written

$$\int f(x)\,dx = F(x) + C$$

Since the derivative of any constant is zero, $x^2 + 1$, $x^2 - 7$ and $x^2$ all differentiate to $2x$. Reversing the process cannot recover which constant was there, so every indefinite integral carries $+C$.

Indefinite integrals

$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \ne -1)$$

Raise the power by one and divide by the new power — exactly the reverse of the differentiation rule.

$$\int x^3\,dx = \frac{x^4}{4} + C$$ $$\int (3x^2 - 4x + 5)\,dx = x^3 - 2x^2 + 5x + C$$

Reversing the chain rule

$$\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$$

The extra division by $a$ compensates for the factor the chain rule would introduce.

$$\int (2x+1)^5\,dx = \frac{(2x+1)^6}{12} + C$$

Check by differentiating: $\dfrac{6(2x+1)^5 \times 2}{12} = (2x+1)^5$ ✓

Definite integrals

$$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$

The constant cancels in the subtraction, so definite integrals need no $+C$.

$$\int_1^3 2x\,dx = \Big[x^2\Big]_1^3 = 9 - 1 = 8$$

$$\int_0^2 (x^2+1)\,dx = \left[\frac{x^3}{3} + x\right]_0^2 = \left(\frac83 + 2\right) - 0 = \frac{14}{3}$$

Two properties worth knowing:

$$\int_a^b f = -\int_b^a f \qquad \int_a^b f + \int_b^c f = \int_a^c f$$

Area under a curve

For a curve above the $x$-axis between $x = a$ and $x = b$:

$$\text{Area} = \int_a^b y\,dx$$

Find the area under $y = x^2$ from $x = 0$ to $x = 3$.

$$\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = 9 - 0 = 9$$

For an area measured from the $y$-axis, integrate with respect to $y$ instead:

$$\text{Area} = \int_c^d x\,dy$$

Area between curves

$$\text{Area} = \int_a^b (y_{\text{upper}} - y_{\text{lower}})\,dx$$

where $a$ and $b$ are the $x$-coordinates of the intersection points.

Find the area between $y = x$ and $y = x^2$.

They meet where $x = x^2$, so $x = 0$ and $x = 1$. Between those, $x > x^2$, so the line is on top:

$$\int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac12 - \frac13 = \frac16$$

Volumes of revolution

Rotating the region under $y = f(x)$ about the $x$-axis produces a solid of volume

$$V = \pi\int_a^b y^2\,dx$$

Rotate $y = x$ from $x = 0$ to $x = 2$ about the $x$-axis.

$$V = \pi\int_0^2 x^2\,dx = \pi\left[\frac{x^3}{3}\right]_0^2 = \frac{8\pi}{3}$$

This is a cone of radius 2 and height 2, and the standard formula gives $\frac13\pi r^2 h = \frac13\pi(4)(2) = \frac{8\pi}{3}$ ✓

About the $y$-axis instead:

$$V = \pi\int_c^d x^2\,dy$$

Integrating trigonometric functions

Reading the derivatives backwards:

$$\int \sin x\,dx = -\cos x + C$$ $$\int \cos x\,dx = \sin x + C$$ $$\int \sec^2 x\,dx = \tan x + C$$

$$\int_0^{\pi/2}\cos x\,dx = \Big[\sin x\Big]_0^{\pi/2} = 1 - 0 = 1$$

With a linear inside function, divide by its coefficient:

$$\int \sin(3x)\,dx = -\frac{\cos(3x)}{3} + C$$

Check: $\dfrac{d}{dx}\left(-\dfrac{\cos 3x}{3}\right) = \dfrac{3\sin 3x}{3} = \sin 3x$ ✓