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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #14 : Integrals

If $f(x)=\cos x$ then

$f^{'}\left ( \frac{\pi }{2} \right )=$

Possible Answers:

$\frac{2^{1/2}}{-2}$

$\frac{1}{2}$

Correct answer:

Explanation:

The answer is .

f(x)=cos(x)

f'(x)=-sin(x)

We know that

$\frac{\pi }{2}=90^{\circ}$

so,

$f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1$

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Example Question #7 : Numerical Approximations To Definite Integrals

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=$

Possible Answers:

1/2

1/4

does not exist

Correct answer:

Explanation:

When we let x = 0 in our original limit, we obtain the 0/0 indeterminate form. Therefore, we can apply L'Hospital's Rule, which requires that we take the derivative of the numerator and denominator separately.

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}2\sin(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}x\tan(x)}$

Apply the Chain Rule in the numerator and the Product Rule in the denominator.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}$

If we again substitute x = 0, we still obtain the 0/0 indeterminate form. Thus, we can apply L'Hospital's Rule one more time.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}4x\cos(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x)+x\sec^2(x))}$

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}$

If we now let x = 0, we can evaluate the limit.

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}=\frac{4\cos(0)+0}{\sec^2(0)+\sec^2(0)+0}$

$=\frac{4}{1+1}=2$

The answer is 2.

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Example Question #11 : Numerical Approximations To Definite Integrals

Differentiate $y=3\csc x+5\sin x$

Possible Answers:

$3\csc x\cot x-5\cos x$

$-3\csc x\cot x+5\cos x$

$3csc^{2}x-5cosx$

$3sec^{2}x-5cosx$

$\cot x(3\csc x-5)$

Correct answer:

$-3\csc x\cot x+5\cos x$

Explanation:

The answer is $-3\csc x\cot x+5\cos x$

We simply differentiate by parts, remembering our trig rules.

$y=3\csc x+5\sin x$

$y^{'}= -3\csc x\cot x+5\cos x$

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Example Question #14 : Integrals

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find h'(x) .

h(x)=fg

Possible Answers:

Correct answer:

Explanation:

The answer is 10.

h(x)=fg

h'(x)=f'(x)g(x)+f(x)g'(x)

h'(x)=(2)(3)+(1)(4) =10

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Example Question #231 : Ap Calculus Ab

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

Possible Answers:

$h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

$h'(x)=\ln (\frac{1}{x})\cdot x^{\frac{1}{x}-1}$

$h'(x)=x^{\frac{1}{x}}\ln (x^{2}-1)$

$h'(x)=\frac{1-\ln x}{x^{2}}$

$h'(x)=\frac{1}{x}\cdot x^{\frac{1}{x}-1}$

Correct answer:

$h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Explanation:

First, let y=h(x) .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let y=h(x) , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with h'(x) .

The answer is $h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$ .

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Example Question #12 : Numerical Approximations To Definite Integrals

Consider the curve given by the parametric equations below:

$x(t)=\ln(t)$

y(t)=3^t-2^t

What is the equation of the line normal to the curve when t=1 ?

Possible Answers:

$y=\frac{-1}{\ln \frac{27}{4}}x-1$

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

$y=\frac{-1}{\ln \frac{27}{4}}(x-1)$

$y-1={\ln \frac{27}{4}}(x+1)$

$y=\frac{1}{\ln \frac{4}{27}}x+1$

Correct answer:

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

Explanation:

In order to find the equation of the normal line, we will need the slope of the line and a point through which it passes. If we substitute t=1 into our parametric equations, we can easily obtain the point on the curve.

x(1)=0, y(1)=1

The normal line is perpendicular to the tangent line. Thus, we should first find the slope of the tangent line.

To find the value of the tangent slope when t=1 , we will use the following formula:

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$

$y'(t)=3^t \ln 3 -2^t \ln 2$

$y'(1)=3\ln3-2\ln 2=\ln \frac{27}{4}$

$x'(t)=\frac{1}{t}$

x'(1)=1

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\ln(\frac{27}{4})$

Because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

slope of normal = $\frac{-1}{\ln(\frac{27}{4})}$ .

We now have the point and slope of the normal line, so we can use point-slope form.

$y-1=\frac{-1}{\ln \frac{27}{4}}(x-0)$

The answer is $y=\frac{-1}{\ln \frac{27}{4}}x+1$ .

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Example Question #21 : Functions, Graphs, And Limits

Evaluate the definite integral of the algebraic function.

integral (x³+ √(x))dx from 0 to 1

Possible Answers:

5/12

10/12

11/12

Correct answer:

11/12

Explanation:

Step 1: Rewrite the problem.

integral (x³+x^1/2)dx from 0 to 1

Step 2: Integrate

x⁴/4 + 2x^(2/3)/3 from 0 to 1

Step 3: Plug in bounds and solve.

[1⁴/4 + 2(1)^(2/3)/3] – [0⁴/4 + 2(0)^(2/3)/3] = (1/4) + (2/3) = (3/12) + (8/12) = 11/12

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Example Question #22 : Functions, Graphs, And Limits

Evaluate the integral.

Integral from 1 to 2 of (1/x³) dx

Possible Answers:

–3/8

–5/8

3/8

1/2

Correct answer:

3/8

Explanation:

Integral from 1 to 2 of (1/x³) dx

Integral from 1 to 2 of (x^-3) dx

Integrate the integral.

from 1 to 2 of (x^–2/-2)

(2^–2/–2) – (1^–2/–2) = (–1/8) – (–1/2)=(3/8)

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Example Question #232 : Ap Calculus Ab

Evaluate the following indefinite integral.

$\int 4 dx$

Possible Answers:

4x^2

x+C

4x+C

Correct answer:

4x+C

Explanation:

In order to evaluate the indefinite integral, ask yourself, "what expression do I differentiate to get 4". Next, use the power rule and increase the power of by 1. To start, we have $4x^{0}$ , so in the answer we have $4x^{1}$ . Next add a constant that would be lost in the differentiation. To check your work, differentiate your answer and see that it matches "4".

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Example Question #2 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate the following indefinite integral.

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

Possible Answers:

4x^4-3x^3+3x+C

x^4-x^3+3x+C

$\frac{1}{4}x^4-\frac{1}{3}x^3+3x+C$

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

$\frac{1}{4}x^4-\frac{1}{3}x^3+3x$

Correct answer:

$\frac{1}{4}x^4-\frac{1}{3}x^3+3x+C$

Explanation:

Use the inverse Power Rule to evaluate the integral. We know that $\int x^n dx= \frac{x^{n+1}}{n+1}+C$ for $n \ne -1$ . We see that this rule tells us to increase the power of by 1 and multiply by $\frac{1}{n+1}$ . Next always add your constant of integration that would be lost in the differentiation. Take the derivative of your answer to check your work.

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #14 : Integrals

Example Question #7 : Numerical Approximations To Definite Integrals

Example Question #11 : Numerical Approximations To Definite Integrals

Example Question #14 : Integrals

Example Question #231 : Ap Calculus Ab

Example Question #12 : Numerical Approximations To Definite Integrals

Example Question #21 : Functions, Graphs, And Limits

Example Question #22 : Functions, Graphs, And Limits

Example Question #232 : Ap Calculus Ab

Example Question #2 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

All AP Calculus AB Resources

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