Integrals - AP Calculus AB

Card 0 of 3487

Question

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

h(t)=-15t^2+30t feet.

At what time will the velocity equal zero?

Answer

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

Compare your answer with the correct one above

Question

Write the domain of the function.

$f(x)=\frac{(x^{4}-81)^{1/2}}{x-4}$

Answer

The answer is ${x\mid x eq 4\ and\ \left | x \right |\geq 3}$

The denominator must not equal zero and anything under a radical must be a nonnegative number.

Compare your answer with the correct one above

Question

Find $\lim _{x\rightarrow 0}f(x)$

Answer

The one side limits are not equal: left is 0 and right is 3

Compare your answer with the correct one above

Question

Which of the following is a vertical asymptote?

Answer

When approaches 3, approaches $\pm \infty$ .

Vertical asymptotes occur at values. The horizontal asymptote occurs at

.

Compare your answer with the correct one above

Question

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$

Answer

First, let's multiply the numerator and denominator of the fraction in the limit by $\frac{1}{x^{4}}$ .

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$ $=\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}$

$=\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}$

As becomes increasingly large the $\frac{1}{x^{4}}$ and $\frac{1}{x^{2}} ^{}$ terms will tend to zero. This leaves us with the limit of $\frac{1}{4}$ .

$\lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}$ .

The answer is $\frac{1}{4}$ .

Compare your answer with the correct one above

Question

Let and be inverse functions, and let

$\f(3)=4 \g(3)=-5 \g(4)=3 \f'(3)=-2 \f'(4)=4 \g'(3)=-1$ .

What is the value of ?

Answer

Since and are inverse functions, . We can differentiate both sides of the equation with respect to to obtain the following:

$g'(f(x))\cdot f'(x)=1$

We are asked to find , which means that we will need to find such that . The given information tells us that , which means that . Thus, we will substitute 3 into the equation.

$g'(f(3))\cdot f'(3)=1$

The given information tells us that.

The equation then becomes $g'(4)\cdot (-2)=1$ .

We can now solve for .

$g'(4)=-\frac{1}{2}$ .

The answer is $-\frac{1}{2}$ .

Compare your answer with the correct one above

Question

Using the fundamental theorem of calculus, find the integral of the function from to .

Answer

The fundamental theorem of calculus is, $\int_{a}^{b} f(x)\ dx=F(b)-F(a)$ , now lets apply this to our situation.

$\int_{0}^{1} x^3+2x+9\ dx$

We can use the inverse power rule to solve the integral, which is $\int x^n\ dx =\frac{x^{n+1}}{n+1}+C$ .

$\int_{0}^{1} x^3+2x+9\ dx=\frac{x^4}{4}+x^2+9x \Big|^{1}_{0}$

$=\left[ \frac{1^4}{4}+1^2+9(1)\right]-\left[\frac{0^4}{4}+0^2+9(0) \right ]$

$=\frac{1}{4}+1+9-0=\frac{41}{4}$

Compare your answer with the correct one above

Question

$f(x) = \frac{x+1}{\sqrt{x^2 -4}}$

What are the horizontal asymptotes of ?

Answer

Compute the limits of as approaches infinity.

Compare your answer with the correct one above

Question

What is the value of the derivative of $f(x)=x^3+6x^2+\frac{3}{4}x$ at x=1?

Answer

First, find the derivative of the function, which is:

$f{}'(x)=3x^2+12x+\frac{3}{4}$

Then, plug in 1 for x:

$3(1^2)+12(1)+\frac{3}{4}$

The result is $\frac{63}{4}$ .

Compare your answer with the correct one above

Question

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

y = ln(x2) at (e, 3)

Answer

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)

Compare your answer with the correct one above

Question

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

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

Answer

The answer is 1.

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

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

$h'(x)=\frac{2(32-41)}{4} = 1$

Compare your answer with the correct one above

Question

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

Answer

The answer is 10.

Compare your answer with the correct one above

Question

Find the equation of the tangent line at on graph

Answer

The answer is

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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.)

Compare your answer with the correct one above

Question

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

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

Answer

The answer is .

We know that

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

so,

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

Compare your answer with the correct one above

Question

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

Answer

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

We simply differentiate by parts, remembering our trig rules.

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

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

Compare your answer with the correct one above

Question

Find the equation of the tangent line at when

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

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.

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

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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 $-\infty$ .

Compare your answer with the correct one above

Question

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

Answer

First, let .

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

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 , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with .

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

Compare your answer with the correct one above

Tap the card to reveal the answer