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

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Example Question #1 : Calculus I — Derivatives

Consider the function

f(x)=x^2-x

Find the minimum of the function on the interval $\left [ 0,1 \right ]$ .

Possible Answers:

-0.5

0.3

-0.25

Correct answer:

-0.25

Explanation:

To find potential minima of the function, take the first derivative of f(x) using the power rule.

f'(x) = 2x-1

Set the derivative to 0:

0 = 2x-1

We solve for to obtain x=0.5 and then plug in 0.5 into the original function to obtain the answer of f(0.5)=-.25

We can double check that x=0.5 is indeed a minimum by using the second derivative test

f''(x) = 2 > 0

which means the function is concave up, so that the point we found is a minimum.

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Example Question #1 : Techniques Of Antidifferentiation

Find the x coordinate of the minimum of f(x)=3x^3-2x^2 on the interval $\left [ -2, 2\right ]$ .

Possible Answers:

$\frac{4}{9}$

Correct answer:

Explanation:

First, find the derivative, which is:

9x^2-4x

Set that equal to 0 to find your critical points:

$9x^2-4x=0; x=0, \frac{4}{9}$

Evaluate f at each critical point and endpoint over [-2, 2]:

f(-2)=-32

f(0)=0

$f(\frac{4}{9})=-\frac{32}{243}$

f(2)=16

The least of those numbers is the minimum. Because f(-2)=-32, that is the minimum.

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Example Question #1 : Calculus I — Derivatives

What is the local minimum of x^3+2x^2+5 when $-1\leq x\leq 1$ ?

Possible Answers:

x=-1

x=0

There is no local minimum in that range.

The y-value is constant throughout that range.

x=1

Correct answer:

x=0

Explanation:

To find the maximum, we need to look at the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as 5x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{2})+(2*2x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that 3x^2+4x has a root when x=0 . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval $-1 \leq x \leq 1$ .

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Example Question #2 : Techniques Of Antidifferentiation

Integrate,

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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Example Question #1 : Techniques Of Antidifferentiation

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

Possible Answers:

-1.61

-8.40

-13.08

-0.41

Correct answer:

-1.61

Explanation:

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

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Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{27}^{31}(\frac{(8x)}{51})dx\end{align*}$

Possible Answers:

18.20

7.00

100.08

2.19

Correct answer:

18.20

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4x^{2})}{51}|_{27}^{31}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4(31)^{2})}{51}-(\frac{(4(27)^{2})}{51})\\&\int_{27}^{31}(\frac{(8x)}{51})dx=18.20\end{align*}$

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Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx\end{align*}$

Possible Answers:

0.64

0.34

1.99

0.24

Correct answer:

1.99

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{x}{4})})}{(7ln(2))}|_{-6}^{-3}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{(-3)}{4})})}{(7ln(2))}-(\frac{(40\cdot 2^{(\frac{(-6)}{4})})}{(7ln(2))})\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=1.99\end{align*}$

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Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx\end{align*}$

Possible Answers:

-5.21

-1.53

-0.16

-10.41

Correct answer:

-1.53

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln(x))}{9}|_{-24}^{-19}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln((-19)))}{9}-(\frac{(59ln((-24)))}{9})\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=-1.53\end{align*}$

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Example Question #4 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-6}^{-2}(\frac{11}{59})dx\end{align*}$

Possible Answers:

6.71

0.20

0.11

0.75

Correct answer:

0.75

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11x)}{59}|_{-6}^{-2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11(-2))}{59}-(\frac{(11(-6))}{59})\\&\int_{-6}^{-2}(\frac{11}{59})dx=0.75\end{align*}$

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Example Question #5 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx\end{align*}$

Possible Answers:

0.15

8.25

3.50

1.25

Correct answer:

1.25

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin(x + 2))}{11}|_{-16}^{-14}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin((-14) + 2))}{11}-(\frac{(9sin((-16) + 2))}{11})\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=1.25\end{align*}$

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

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Calculus I — Derivatives

Example Question #1 : Techniques Of Antidifferentiation

Example Question #1 : Calculus I — Derivatives

Example Question #2 : Techniques Of Antidifferentiation

Example Question #1 : Techniques Of Antidifferentiation

Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #4 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #5 : Antiderivatives Following Directly From Derivatives Of Basic Functions

All AP Calculus AB Resources

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