We start with the first rule of integration, which allows us to start calculating explicit integrals.
Definition
One of the most fundamental rules in differential calculus is the power rule,
\frac{d}{dx} x^m = mx^{m-1} .
Suspecting this will be a useful property, we can derive an expression for the inverse power rule from the definition of an integral.
\int mx^{m-1} dx = x^m
Applying linearity and the substitution n = m-1, we obtain
\int x^{n} dx = \frac{x^{n+1}}{n+1}
whenever n \not = -1. This is known as the inverse power rule, and allows us to integrate any regular polynomial.
For every n \in \mathbb{R} \backslash \{-1\}
\int x^{n} dx = \frac{x^{n+1}}{n+1} + C.
This is the inverse power rule.
Lets apply this rule to some polynomials.
Find the solution to the following integral
\int 2x^2 - 3x - 1 \, dx
We first begin by applying linearity
= 2 \int x^2 \,dx - 3\int x \, dx - \int 1x^0 \, dx
Then inverse power rule
= \frac{2x^3}{3} - \frac{3x^2}{2} - x + C
This process can be done for any standard polynomial.
Let f(x) be a polynomial in the form
f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + c
Then
\int f(x)\,dx = \frac{a_nx^{n+1}}{n+1} + \frac{a_{n-1}x^{n}}{n} + \cdots + \frac{a_1x^2}{2} + cx + C
But that’s not all! The inverse power rule works for all exponents (except -1) so we can apply this rule to roots and such.
Find the solution to the following integral
\int \frac{x}{\sqrt[3]{x}} \, dx
Changing exponents allows us to get the following
\int \frac{x}{\sqrt[3]{x}} \, dx = \int x^{2/3} \, dx
Then inverse power rule
= \frac{3x^{5/3}}{5} + C
This also works for real exponents, but it’s not very useful in practice.
Problems
These problems will only cover inverse power rule, and linearity.
Integrals involving Integer exponents
Evaluate the following integral
\int x \, dx
Evaluate the following integral
\int x^2 \, dx
Evaluate the following integral
\int 2x^2 - 3x - 1 \, dx
Evaluate the following integral
\int 2x(x^2 + 4)^2 \, dx
Evaluate the following integral
\int \frac{1}{x^2} \, dx
Evaluate the following integral
\int \frac{2}{x^6} + \frac{4}{x^{10}} - \frac{8}{x^{12}} \, dx
Integrals involving Fractional exponents
Evaluate the following integral
\int \sqrt{x} \, dx
Evaluate the following integral
\int \frac{1}{\sqrt{x}} \, dx
Evaluate the following integral
\int \sqrt[3]{x} + 2\sqrt[5]{x^2}\, dw
Special problems
Evaluate the following integral
\int \, dx
Evaluate the following integral
\int x^\pi \, dx
Evaluate the following integral
\int 6x^5\, dx - 18x^2 + 7
Notice the integral ends at the dx, meaning the solution is
\int 6x^5 - 18x^2 + 7 \, dx = x^6 - 18x^2 + C
Evaluate the following integral
\int 12t^7 - t^2 - t + 3 \, dt
Classic polynomial, the only difference is the use of t instead of x
\int 12t^7 - t^2 - t + 3 \, dt = \frac{3}{2}t^8 - \frac{1}{3}t^3 - \frac{1}{2}t^2 + 3t + C
Evaluate the following integral
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw
Convert the radicals to fractional exponents
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \int w^\frac{1}{3} + 10w^\frac{3}{5} \, dw
And then apply inverse power rule
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \frac{3}{4}w^\frac{4}{3} + \frac{25}{4}w^\frac{8}{5} + C
Evaluate the following integral
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy
Convert the radicals to fractional exponents
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = \int \frac{7}{3}y^{-6} + y^{-10} - 2y^{\frac{4}{3}} \, dy
And then apply inverse power rule
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = -\frac{7}{15}y^{-5} - \frac{1}{9} y^{-9} + 6y^{-\frac{1}{3}} + C