AP Calculus AB - Antiderivatives following directly from derivatives of basic functions

Find the antiderivative of the following.

$\sec^2(x)$

$\tan (x)$

$2\sec (x)$

$\frac{\tan (x)}{2}$

$2\tan (x)$

Explanation

$\sec^2(x)$ is the derivative of $\tan (x)$ . Thus, the antiderivative of $\sec^2(x)$ is $\tan (x)$ .

Given , find the general form for the antiderivative .

$\frac{4}{3}x^3 +10x^2 +25x+C$

None of the other answers

Explanation

To answer this, we will need to FOIL our function first.

Now can find the antiderivatives of each of these three summands using the power rule.

$f(x) = \frac{4}{3}x^3 +10x^2 +25x+C$ (Don't forget )!

Determine the value of $\int_{0}^{2}\sqrt{9x^2+12x+4} dx$ .

Explanation

We can factor the equation inside the square root:

$\\int_0^2\sqrt{9x^2+12x+4}dx \\\int_0^2\sqrt{(3x+2)^2}dx \\\int_0^2 (3x+2)dx$

From here, increase each term's exponent by one and divide the term by the new exponent.

$\frac{3x^2}{2}+2x\big|_0^2$

Now, substitute in the upper bound into the function and subtract the lower bound function value from it.

$\\frac{3(2)^2}{2}+2(2)-\left(\frac{3(0)^2}{2}+2(0)\right) \\\frac{3(4)}{2}+4 \\\frac{12}{2}+4 \\6+4 \\10$

Therefore,

$\int_{0}^{2}\sqrt{(3x+2)^2}dx=\int_{0}^{2}(3x+2)dx=10$

Integrate,

$\int[x^2 +\sin(x)-3]dx$

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

$\frac{1}{2}x^2 + \cos(x) - 3x +C$

$x -x^2+ \cos(x) +C$

$\frac{1}{2}x - \sin(x) - 3x^2 +C$

$\frac{1}{2}x^3 -\cos(x) - \frac{3}{2}x^2 +C$

Explanation

Integrate

$\int[x^2 +\sin(x)-3]dx$

1) Apply the sum rule for integration,

$\int x^2 dx +\int \sin(x)dx -\int 3dx$

2) Integrate each individual term and include a constant of integration,

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

$\frac{d}{dx}\left(\frac{1}{3}x^3-\cos(x)-3x+C \right )=x^2+\sin(x) -3$

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.

$\begin{align*}&\text{Evaluate the integral: }\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx\end{align*}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin(x + 1))}{8}|_{8}^{11.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin((11.5) + 1))}{8}-(\frac{(27sin((8) + 1))}{8})\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=-1.61\end{align*}$

Integrate:

$\int (x^2 +\frac{1}{2x})dx$

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |$

$\frac{x^3}{3}+\frac{2}{\ln(x)}+C$

$x^3+\frac{1}{2}\ln\left | x\right |+C$

Explanation

To evaluate the integral, we can split it into two integrals:

$\int x^2 dx+ \frac{1}{2}\int \frac{dx}{x}$

After integrating, we get

$\frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C$

where a single constant of integration comes from the sum of the two integration constants from the two individual integrals, added together.

The rules used to integrate are

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \frac{dx}{x}=\ln\left | x\right | +C$

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

$-4cos(t)+12sin(t)+\frac{t^3}{3}+c$

Explanation

Calculate the following integral.

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called antiderivatives. This means that we can calculate integrals by reversing our integration rules.

Thus, we can have the following rules.

$\int sin(t)dt = -cos(t)+c$