AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

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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 – 3 = (2/e)(x – e)

y – 3 = ln(e2)(x – e)

y = (2/e)(x – e)

y – 3 = (x – e)

y = (2/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)

Example Question #1 : Numerical Approximations To Definite Integrals

Find the equation of the tangent line at  when

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

Possible Answers:

Correct answer:

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.

Example Question #1 : Numerical Approximations To Definite Integrals

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

Possible Answers:

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

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

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

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

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

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}}

Example Question #2 : Numerical Approximations To Definite Integrals

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

Possible Answers:

1

\infty

0

-2

-\infty

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.

Example Question #11 : Integrals

Evaluate:

Possible Answers:

cannot be determined

Correct answer:

Explanation:

First, we can write out the first few terms of the sequence , where  ranges from 1 to 3.

Notice that each term , is found by multiplying the previous term by . Therefore, this sequence is a geometric sequence with a common ratio of . We can find the sum of the terms in an infinite geometric sequence, provided that , where  is the common ratio between the terms. Because  in this problem,  is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

The answer is .

 

Example Question #1 : Numerical Approximations To Definite Integrals

If  then find .

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

Example Question #11 : Integrals

Find the equation of the tangent line at  on graph 

Possible Answers:

Correct answer:

Explanation:

The answer is 

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

f'(x)=2x+2

 

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

 

Example Question #1 : Numerical Approximations To Definite Integrals

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

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

Possible Answers:

Correct answer:

Explanation:

The answer is 

 

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.)

 

Example Question #2 : Numerical Approximations To Definite Integrals

If  then 

Possible Answers:

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

\frac{1}{2}

Correct answer:

Explanation:

The answer is .

 

 

We know that

so,

 

Example Question #1 : Numerical Approximations To Definite Integrals

Possible Answers:

1/4

1/2

1

does not exist

2

Correct answer:

2

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.

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

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.

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

The answer is 2.

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