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AP Calculus AB : Numerical approximations to definite integrals

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Example Questions

Example Question #1 : Integrals

Write the equation of a tangent line to the given function at the point.

y = ln(x²) at (e, 3)

Possible Answers:

y = (2/e)(x – e)

y = (2/e)

y – 3 = ln(e²)(x – e)

y – 3 = (x – e)

y – 3 = (2/e)(x – e)

Correct answer:

y – 3 = (2/e)(x – e)

Explanation:

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x²)

y' = (2x/(x²))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e²)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Example Question #1 : Numerical Approximations To Definite Integrals

Find the equation of the tangent line at (1,-1) when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

Possible Answers:

y=x+1

y=1

y=x-2

y=-x-2

y=x

Correct answer:

y=x-2

Explanation:

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

y+1=1(x-1)

y+1=x-1

y=x-2

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Example Question #11 : Integrals

Differentiate $y=\frac{t+2}{t^{2}-4t-12}$

Possible Answers:

$y'=\frac{1}{(t^{2}-4t-12)^{2}}$

$y'=\frac{1}{(t-6)}$

$y'=\frac{(t+2)-1}{(t^{2}-4t-12)^{2}}$

$y'=\frac{-1}{(t-6)^{2}}$

$y'=\frac{1}{2t-4}$

Correct answer:

$y'=\frac{-1}{(t-6)^{2}}$

Explanation:

We see the answer is $y'=\frac{-1}{(t-6)^{2}}$ after we simplify and use the quotient rule.

$y=\frac{t+2}{t^{2}-4t-12}$ we could use the quotient rule immediatly but it is easier if we simplify first.

$y=\frac{t+2}{(t+2)(t-6)}$

$y=\frac{1}{(t-6)}$

$y=(t-6)^{-1}$

$y'=-(t-6)^{-2}$

$y'=\frac{-1}{(t-6)^{2}}$

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Example Question #11 : Integrals

Find $\lim_{x\rightarrow \infty }\frac{-2x^3+x^2-2}{x^2+10}$

Possible Answers:

$1$

$0$

$\infty$

$-\infty$

$-2$

Correct answer:

$-\infty$

Explanation:

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has $-2x^3$ and the denominator has $x^2$ . Therefore, by cancellation, it becomes $-2x$ as approaches infinity. So the answer is $-\infty$ .

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Example Question #12 : Integrals

Evaluate:

$\sum_{n=1}^{\infty }4(\frac{-1}{2})^{n-1}$

Possible Answers:

cannot be determined

$\infty$

$\frac{8}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

First, we can write out the first few terms of the sequence $a_{n}=$ $4(\frac{-1}{2})^{n-1}$ , where ranges from 1 to 3.

$a_{1}=4(\frac{-1}{2})^{1-1}=4$

$a_{2}=4(\frac{-1}{2})^{2-1}=-2$

$a_{3}=4(\frac{-1}{2})^{3-1}=1$

Notice that each term $\left ( n>1 \right )$ , is found by multiplying the previous term by $-\frac{1}{2}$ . Therefore, this sequence is a geometric sequence with a common ratio of $-\frac{1}{2}$ . We can find the sum of the terms in an infinite geometric sequence, provided that |r|<1 , where is the common ratio between the terms. Because $r=-\frac{1}{2}$ in this problem, $\left | r \right |$ is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

$S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}$

The answer is $\frac{8}{3}$ .

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Example Question #11 : Integrals

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

$h(x)=\frac{2f}{g}$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(3*2-4*1)}{4} = 1$

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Example Question #2 : Numerical Approximations To Definite Integrals

Find the equation of the tangent line at (4,1) on graph f(x)=(x-2)(x+4)

Possible Answers:

y=3x-7

y=7x+8

y=x-11

y=x+40

y=10x-39

Correct answer:

y=10x-39

Explanation:

The answer is 10x-39

f(x)=(x-2)(x+4)

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

f'(x) = 2(4)+2=10

(This is the slope. Now use the point-slope formula)

y-1=10(x-4)

y-1=10x-40

y=10x-39

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Example Question #13 : Integrals

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

Possible Answers:

y=2ax-b+1

y=2ab+b-1

y=2ax+bx-2a-b+1

y=2ax-bx-2a+1

y=2a+b

Correct answer:

y=2ax+bx-2a-b+1

Explanation:

The answer is y=2ax+bx-2a-b+1

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

y-1=(2a+b)(x-1)

y-1=2ax+bx-2a-b

y=2ax+bx-2a-b+1

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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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AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Integrals

Example Question #1 : Numerical Approximations To Definite Integrals

Example Question #11 : Integrals

Example Question #11 : Integrals

Example Question #12 : Integrals

Example Question #11 : Integrals

Example Question #2 : Numerical Approximations To Definite Integrals

Example Question #13 : Integrals

Example Question #14 : Integrals

Example Question #7 : Numerical Approximations To Definite Integrals

All AP Calculus AB Resources

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