A function F(x) is called an antiderivative of f(x) when:
F'(x) = f(x) \implies \frac{d F(x)}{dx} = f(x) \implies F(x) = \int f(x) dx
| f(x) | \int f(x) dx |
|---|---|
| a | ax + C |
| x^n | \frac{x^{n+1}}{n+1} + C, a \ne -1 |
| \frac{1}{x} | \ln\vert x \vert + C |
| e^x | e^x + C |
| a^x | \frac{a^x}{\ln(a)} + C |
| \vert x \vert | \frac{x\vert x \vert}{2} + C |
| \ln(x) | x\ln(x) - x + C |
| \sin(x) | -\cos(x) + C |
| \cos(x) | \sin(x) + C |
| \frac{1}{a^2 + x^2} | \frac{1}{a}\arctan(\frac{x}{a}) + C |
| \frac{1}{\sqrt{a^2 - x^2}} | \arcsin\frac{x}{a} + C |
| \sqrt{a^2 - x^2} | \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \arcsin\frac{x}{a} + C |
| \frac{1}{\sqrt{a^2 + x^2}} | \ln(x + \sqrt{a^2 + x^2}) + C |
| \sqrt{a^2 + x^2} | \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln(x + \sqrt{a^2 + x^2}) + C |
\int af(x) dx = a \int f(x) dx
\int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx
Let y = g(x)
\int f(g(x))g'(x) dx = \int f(y)dy
\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx
In other words:
\int fdg = fg - \int gdf
We need to always add a constant after integration because it is lost during derivation:
Let f(x) = x + 1 then f'(x) = 1, then we integrate \int f'(x) dx = x. As we can see, the +\ 1 was lost.
Functions differing by a constant are called a family of primitive functions.
To find \int \frac{f(x)}{g(x)} dx two things have to be satisfied:
Then the following steps can be taken:
Taking an integral of the partial fractions is much simpler.
Let t = \tan\frac{x}{2} then trigonometrical functions can be reduced to a rational function:
dx is then replaced with \frac{2}{1 + t^2}dt
Powerful tool for finding areas under graphs
Let f be continuous on some closed interval [a,b]. Now we evenly divide it into n subintervals. If P is the partition of all edges of intervals then P = \{ x_0 = a, x_1, \cdots, x_{n-1}, x_n = b\}. To find the area of some subinterval of f at some point c_k: f(c_k) \cdot (x_k - x_{k-1}) = f(c_k) \cdot \Delta_{x_k}. Therefore to find the area of the whole interval:
\sum_{k=1}^n f(c_k) \cdot \Delta_{x_k}
To find the area (I) of some f on a closed interval [a,b] n has to approach infinity
\lim_{n \to \infty} \sum_{k=1}^n f(c_k) \cdot \Delta_{x_k} = I
I is called the definite integral of f over [a,b]
I = \int_a^bf(x)dx = F(b) - F(a)
where F' = f
\frac{d}{dx} \int_a^{g(x)} f(t)dt = f(g(x))g'(x)
\frac{d}{dx} \int_a^b f(x)dx = 0
Same as with indefinite integrals plus:
\int_a^b f(g(x)) \cdot g'(x) dx = \int_{g(a)}^{g(b)} f(u) du
While this is true, finding c is incredibly hard:
\int_a^b f(x) dx = (b-a)f(c)
A function with a continuous first derivative is called smooth and its graph is a smooth curve.
If in a range [a,b] f(x) \le g(x) then the area between the curves is
\int_a^b |g(x) - f(x)| dx
Let f be a smooth function. The length of the curve from a to b is then:
\int_a^b \sqrt{1 + (f'(x))^2}dx
If we rotate some region P (marked by the area under the graph of some function f from a to b) about the x axis the formula for its volume is:
\pi \int_a^b (f(x))^2dx
Integrals with infinite limits are called improper integrals.
\int_a^\infty f(x)dx = \lim_{b\to\infty} \int_a^b f(x)dx
If the limit is finite we say the improper integral converges and the limit is the resulting value. Otherwise it diverges.
Just like with series, \int_1^\infty \frac{1}{x^p}dx converges only if p > 1, converges to \frac{1}{p-1}
\int_{-\infty}^\infty f(x)dx = \int_{-\infty}^c f(x)dx + \int_c^\infty f(x)dx
The left side converges only if the right one does as well.
If one (or both) of the limits of an integral are unbounded, we can define an improper integral with a limit directed towards the interval:
let a be unbounded:
\int_a^b f(x)dx = \lim_{c \to a^+} \int_c^b f(x)dx
let b be unbounded:
\int_a^b f(x)dx = \lim_{c \to b^-} \int_a^c f(x)dx
If the function we are integrating over some interval is not continuous, we split the integral into multiple smaller ones to omit the unbounded points and analyze the left and right side separately.
It is important to note it only tells us where something diverges/converges, the converging values might still be different
let 0 \le f(x) \le g(x) for x \ge a
let f and g be positive continuous functions on [a, \infty). If \lim_{x\to\infty}\frac{f(x)}{g(x)} = L, 0 \le L then \int_a^\infty f(x)dx and \int_a^\infty g(x)dx both converge or both diverge.
Let N \ge 1, a_n = f(n) then \sum_{n=1}^\infty a_n and \int_N^\infty f(x)dx both diverge or both converge.
The find a volume under some region R of a function f(x, y) we use double integrals (assuming f(x, y) is continuous over R):
\iint\limits_R f(x, y)dxdy
If R is a rectangular region bounded by a \le x \le b and c \le y \le d, then:
\int_{a}^b \big(\int_{c}^d f(x, y)dy\big)dx = \int_{c}^d \big(\int_{a}^b f(x, y)dx\big)dy
When one component is bounded by a function, it has to be integrated first, example: 0 \le x \le 1 and g(x) \le y \le h(x)
\int_{0}^1 \big(\int_{g(x)}^{h(x)} f(x, y)dy\big)dx
Matrix of all partial derivatives of a function
\mathbf{J} = \begin{bmatrix} \frac{\partial f_{x_1}}{\partial x_1} & \cdots & \frac{\partial f_{x_1}}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_{x_n}}{\partial x_1} & \cdots & \frac{\partial f_{x_n}}{\partial x_n} \\ \end{bmatrix}
The function has to be multiplied by the determinant of the Jacobian matrix when switching from cartesian to polar parametrization.
\iint \limits_{f(A)} f(x, y)dxdy = \iint \limits_{A} f(r\cos\theta, r\sin\theta) \det\mathbf{J} drd\theta
In the case of 2 dimensions for polar \det\mathbf{J} = r, in 3 \det\mathbf{J}_x = r^2\sin\theta
Two variable functions: