# Calculus: Integrals III

## Polynomials

Integrating a polynomial function is as easy as differentiating one:

$$\int x^a \, dx = \dfrac{1}{a+1}x^{a+1} + C, \quad \text{where } a \neq -1$$

Note the exception for when $a=-1$. The integral of $x^{-1}$ is clearly not $\frac{x^0}{0}$. In fact, you may remember that $x^{-1}$ it is the derivative of the natural logarithm, so the natural logarithm is therefore its antiderivative.

Here are some algebraic rules about exponents to keep in mind when integrating polynomials:

• $\dfrac{1}{x^a}=x^{-a}$
• $\sqrt[a]{x}=x^{1/a}$
• $x^{1/a+1} = x^{(a+1)/a} \neq x^{1/(a+1)}$

## Problems

1. Integrate with respect to $x$: $\dfrac{dy}{dx} = x^2$

$\displaystyle\int\dfrac{dy}{dx} \, dx = \displaystyle\int x^2 \, dx\\ y = \dfrac{1}{2+1} x^{2+1} + c\\ y = \dfrac{1}{3} x^{3} + c$

2. Integrate with respect to $x$: $\dfrac{dy}{dx} = \dfrac{1}{3}x^3 - 2x^2 - 2x + 3$

$\displaystyle\int \dfrac{dy}{dx} \, dx = \displaystyle\int \dfrac{1}{3}x^3 - 2x^2 - 2x + 3 \, dx\\ y = \dfrac{1}{3+1}\cdot\dfrac{1}{3}x^{3+1} - \dfrac{1}{2+1}\cdot 2 x^{2+1} -\dfrac{1}{1+1}\cdot 2 x^{1+1} + \dfrac{1}{0+1}3x^{0+1} + c \\ y = \dfrac{1}{4}\cdot\dfrac{1}{3}x^{4} - \dfrac{1}{3}\cdot 2 x^{3} -\dfrac{1}{2}\cdot 2 x^{2} + \dfrac{1}{1}3x^{1} + c \\ y = \dfrac{1}{12}x^{4} - \dfrac{2}{3}x^{3} - x^{2} + 3x + c \\$

3. Evaluate: $\displaystyle\int 7x^3 - 2x + 9\, dx$

$\displaystyle\int 7x^3 - 2x + 9\, dx = 7\left(\dfrac{1}{3+1}\right)x^{3+1} - 2\left(\dfrac{1}{1+1}\right)x^{1+1} + 9x + c \\ \displaystyle\int 7x^3 - 2x + 9\, dx = \dfrac{7}{4}x^{4} - x^{2} + 9x + c \\$

4. Evaluate: $\displaystyle\int 3x^2 + 5x^4 + 6x^6 \, dx$

$\displaystyle\int 3x^2 + 5x^4 + 6x^6 \, dx = 3\left(\dfrac{1}{2+1}x^{2+1}\right) + 5\left(\dfrac{1}{4+1}x^{4+1}\right) + 6\left(\dfrac{1}{6+1}x^{6+1}\right) + c \\ \displaystyle\int 3x^2 + 5x^4 + 6x^6 \, dx = x^{3} + x^{5} + \dfrac{6}{7}x^{7} + c \\$

5. Evaluate: $\displaystyle\int 4x^2 - 4x^{-2} \, dx$

$\displaystyle\int 4x^2 - 4x^{-2} \, dx = \displaystyle\int 4\left(\dfrac{1}{3}x^3\right) - 4\left(\dfrac{1}{-1}x^{-1}\right) +c \\ \displaystyle\int 4x^2 - 4x^{-2} \, dx = \displaystyle\int \dfrac{4}{3}x^3 +4x^{-1} + c \\$

6. Integrate with respect to $x$: $\dfrac{dy}{dx} = \dfrac{1}{x^2}$

$\displaystyle\int \dfrac{dy}{dx} \, dx = \displaystyle\int \dfrac{1}{x^2} \, dx \\ y = \displaystyle\int x^{-2}dx \\ y = \dfrac{1}{-2+1} x^{-2+1} + c\\ y = \dfrac{1}{-1} x^{-1} + c \\ y = \dfrac{-1}{x} + c \\$

7. Evaluate: $\displaystyle\int x^{-1.001} \, dx$

$\displaystyle\int x^{-1.001} \, dx = \dfrac{1}{-1.001+1}x^{-1.001+1} + c \\ \displaystyle\int x^{-1.001} \, dx = \dfrac{1}{-0.001}x^{-0.001} + c\\$

8. Evaluate: $\displaystyle\int \dfrac{1}{x^{-2}} - 4x^{-10} \, dx$

$\displaystyle\int \dfrac{1}{x^{-2}} - 4x^{-10} \, dx = \displaystyle\int x^2 - 4x^{-10} \, dx\\ \displaystyle\int \dfrac{1}{x^{-2}} - 4x^{-10} \, dx = \left(\dfrac{1}{2+1}\right) x^{2+1} - 4\left(\dfrac{1}{-10+1}\right)x^{-10+1} + c \\ \displaystyle\int \dfrac{1}{x^{-2}} - 4x^{-10} \, dx = \dfrac{1}{3}x^{3} + \dfrac{4}{9}x^{-9} + c \\$

9. Integrate with respect to $x$: $\dfrac{dy}{dx} = \left(x+1\right)\left(x^2-1\right)$

$\displaystyle\int \dfrac{dy}{dx} \, dx = \displaystyle\int \left(x+1\right)\left(x^2-1\right) \, dx\\ y = \displaystyle\int x^3+x^2-x-1 \, dx\\ y = \dfrac{1}{3+1}x^{3+1}+\dfrac{1}{2+1}x^{2+1}-\dfrac{1}{1+1}x^{1+1}-1\dfrac{1}{0+1}x^{0+1} +c \\ y = \dfrac{1}{4}x^{4} + \dfrac{1}{3}x^{3} - \dfrac{1}{2}x^{2} - x +c \\$

10. Evaluate: $\displaystyle\int \, dx$

$\displaystyle\int dx = \displaystyle\int 1 \, dx \\ \displaystyle\int \, dx = \int x^{0}\, dx\\ \displaystyle\int \, dx = \dfrac{1}{0+1}x^{0+1} + c\\ \displaystyle\int \, dx = x + c\\$

11. Evaluate: $\int \alpha \, dx$

$\displaystyle\int \alpha dx = \displaystyle\int \alpha x^0 \, dx \\ \displaystyle\int \alpha \, dx = \alpha \displaystyle\int x^0 \, dx \\ \displaystyle\int \alpha \, dx = \alpha\left(\dfrac{1}{0+1}x^{0+1} + c\right)\\ \displaystyle\int \alpha \, dx = \alpha x + c\\$

12. Evaluate: $\displaystyle\int \alpha x + \gamma x^2 - x^{-2} - \omega \, dx$

$\displaystyle\int \alpha x + \gamma x^2 - x^{-2} - \omega \, dx = \displaystyle\int \alpha x + \gamma x^2 - x^{-2} - \omega x^0 \, dx \\ \displaystyle\int \alpha x + \gamma x^2 - x^{-2} - \omega \, dx = \alpha \dfrac{1}{1+1}x^{1+1} + \gamma \dfrac{1}{2+1}x^{2+1} - \dfrac{1}{-2+1}x^{-2+1} - \omega \dfrac{1}{0+1}x^{0+1} + c \\ \displaystyle\int \alpha x + \gamma x^2 - x^{-2} - \omega \, dx = \dfrac{\alpha}{2}x^{2} + \dfrac{\gamma}{3}x^{3} + x^{-1} - \omega x + c \\$

13. Evaluate: $\displaystyle\int 10x^{10} - 2x^e + 8 \, dx$

$\displaystyle\int 10x^{10} - 2x^e + 8 \, dx = 10\left(\dfrac{1}{11}x^{11}\right) - 2\left(\dfrac{1}{e+1}x^{e+1}\right) + 8(x) + c\\ \displaystyle\int 10x^{10} - 2x^e + 8 \, dx = \dfrac{10}{11}x^{11} - \dfrac{2}{e+1}x^{e+1} + 8x + c\\$

14. Evaluate: $\displaystyle\int 2\pi x^{\pi^2} + (\pi+2)x^{(\pi+2)} \, dx$

$\displaystyle\int 2\pi x^{\pi^2} + (\pi+2)x^{(\pi+2)} \, dx = 2\pi\left(\dfrac{1}{\pi^2+1}x^{\pi^2+1}\right) + (\pi+2)\left(\dfrac{1}{\pi+2+1}x^{(\pi+2+1)}\right) + c \\ \displaystyle\int 2\pi x^{\pi^2} + (\pi+2)x^{(\pi+2)} \, dx = \dfrac{2\pi}{\pi^2+1}x^{\pi^2+1} + \dfrac{\pi+2}{\pi+3}x^{(\pi+3)} +c \\$

15. Evaluate: $\displaystyle\int\displaystyle\prod\limits_{i=1}^{66} x^{1.5} \, dx$

$\displaystyle\int \displaystyle\prod\limits_{i=1}^{66} x^{1.5} \, dx = \displaystyle\int \left(x^{1.5}\right)^{66} \, dx \\ \displaystyle\int \displaystyle\prod\limits_{i=1}^{66} x^{1.5} \, dx = \displaystyle\int x^{99} \, dx \\ \displaystyle\int \displaystyle\prod\limits_{i=1}^{66} x^{1.5} \, dx = \dfrac{1}{99+1}x^{99+1} + c \\ \displaystyle\int \displaystyle\prod\limits_{i=1}^{66} x^{1.5} \, dx = \dfrac{1}{100}x^{100} + c \\$

The large capital pi symbol, $\Pi$, is the multiplication analog of the capital sigma symbol, $\Sigma$. To simplify, simple multiply the central term by itself 66 times. In this case, the term is just a monomial in disguise. Had the index variable $i$ been included in the product term, you would have had to account for it as well, just as in sums.