Integrals - AP Calculus AB
Card 0 of 3487
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?
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?
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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.
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.
Write the domain of the function.
$f(x)=\frac{(x^{4}$$$-81)^{1/2}$}{x-4}
Write the domain of the function.
$f(x)=\frac{(x^{4}$$$-81)^{1/2}$}{x-4}
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The answer is 
The denominator must not equal zero and anything under a radical must be a nonnegative number.
The answer is
The denominator must not equal zero and anything under a radical must be a nonnegative number.
Find 
Find
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The one side limits are not equal: left is 0 and right is 3
The one side limits are not equal: left is 0 and right is 3
Which of the following is a vertical asymptote?
Which of the following is a vertical asymptote?
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When 
approaches 3, 
approaches 
.
Vertical asymptotes occur at 
values. The horizontal asymptote occurs at
.
When
Vertical asymptotes occur at
Evaluate the following limit:
$lim_{xrightarrow infty $}$$\frac{1-x^4$$}{x^2$$-4x^4$}$
Evaluate the following limit:
$lim_{xrightarrow infty $}$$\frac{1-x^4$$}{x^2$$-4x^4$}$
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First, let's multiply the numerator and denominator of the fraction in the limit by $$\frac{1}{x^{4}$$}.
$lim_{xrightarrow infty $}$$\frac{1-x^4$$}{x^2$$-4x^$4$}$=lim_{xrightarrow infty $}$$\frac{(frac{1}{x^4$$}$)(1-x^4$$)}{$\frac{1}{x^{4}$$$}(x^2$$-4x^4$)}
$=lim_{xrightarrow 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}$$} ^{} terms will tend to zero. This leaves us with the limit of 
.
$lim_{xrightarrow infty }$$\frac{-1}{-4}$=\frac{1}{4}$.
The answer is $\frac{1}{4}$.
First, let's multiply the numerator and denominator of the fraction in the limit by $$\frac{1}{x^{4}$$}.
$lim_{xrightarrow infty $}$$\frac{1-x^4$$}{x^2$$-4x^$4$}$=lim_{xrightarrow infty $}$$\frac{(frac{1}{x^4$$}$)(1-x^4$$)}{$\frac{1}{x^{4}$$$}(x^2$$-4x^4$)}
$=lim_{xrightarrow infty $}$$\frac{frac{1}{x^4$$}$-1}{$\frac{1}{x^2$}$-4}
As
$lim_{xrightarrow infty }$$\frac{-1}{-4}$=\frac{1}{4}$.
The answer is $\frac{1}{4}$.
Let 
and 
be inverse functions, and let
.
What is the value of 
?
Let
What is the value of
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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}$.
Since
g'(f(x))cdot f'(x)=1
We are asked to find
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}$.
Using the fundamental theorem of calculus, find the integral of the function 
from 
to 
.
Using the fundamental theorem of calculus, find the integral of the function
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The fundamental theorem of calculus is, 
, now lets apply this to our situation.
We can use the inverse power rule to solve the integral, which is 
.
The fundamental theorem of calculus is,
We can use the inverse power rule to solve the integral, which is
What are the horizontal asymptotes of 
?
What are the horizontal asymptotes of
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Compute the limits of 
as 
approaches infinity.
Compute the limits of
What is the value of the derivative of 
at x=1?
What is the value of the derivative of
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First, find the derivative of the function, which is:
Then, plug in 1 for x:
The result is 
.
First, find the derivative of the function, which is:
Then, plug in 1 for x:
The result is
Write the equation of a tangent line to the given function at the point.
y = ln(x2) at (e, 3)
Write the equation of a tangent line to the given function at the point.
y = ln(x2) at (e, 3)
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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)
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)
If 
then find 
.
h(x)=\frac{2f}{g}$
If
h(x)=\frac{2f}{g}$
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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
The answer is 1.
h(x)=\frac{2f}{g}$
$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}$$}
h'(x)=\frac{2(32-41)}{4}$ = 1
If 
then find 
.
If
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The answer is 10.
The answer is 10.
Find the equation of the tangent line at 
on graph 
Find the equation of the tangent line at
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The answer is 
$f(x)=x^{2}$+2x-8
f'(x)=2x+2
(This is the slope. Now use the point-slope formula)
The answer is
$f(x)=x^{2}$+2x-8
f'(x)=2x+2
(This is the slope. Now use the point-slope formula)
Find the equation of the tangent line at (1,1) in
$f(x)=ax^{2}$+bx+c
Find the equation of the tangent line at (1,1) in
$f(x)=ax^{2}$+bx+c
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The answer is 
$f(x)=ax^{2}$+bx+c
f'(x)=2ax+b
f'(1)=2a(1)+b = 2a+b
(This is the slope. Now use the point-slope formula.)
The answer is
$f(x)=ax^{2}$+bx+c
f'(x)=2ax+b
f'(1)=2a(1)+b = 2a+b
(This is the slope. Now use the point-slope formula.)
If 
then
If
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The answer is 
.
We know that
so,
The answer is
We know that
so,
Differentiate 
Differentiate
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The answer is 
We simply differentiate by parts, remembering our trig rules.
The answer is
We simply differentiate by parts, remembering our trig rules.
Find the equation of the tangent line at 
when
$y=\frac{x^{3}$$$-2x(x^{1/2}$)}{x}
Find the equation of the tangent line at
$y=\frac{x^{3}$$$-2x(x^{1/2}$)}{x}
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The answer is
$y=\frac{x^{3}$$$-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}$$)} =1 this is the slope so let's use the point slope formula.
The answer is
$y=\frac{x^{3}$$$-2x(x^{1/2}$)}{x} let's go ahead and cancel out the
$y=x^{2}$$-2(x^{1/2}$)
$y'=2x-$\frac{1}{(x^{1/2}$$)}
$y'(1)=2(1)-$\frac{1}{(1^{1/2}$$)} =1 this is the slope so let's use the point slope formula.
Differentiate $y=\frac{t+2}{t^{2}$$-4t-12}
Differentiate $y=\frac{t+2}{t^{2}$$-4t-12}
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We see the answer is $y'=\frac{-1}{(t-6)^{2}$$} after we simplify and use the quotient rule.
$y=\frac{t+2}{t^{2}$$-4t-12} we could use the quotient rule immediatly but it is easier if we simplify first.
y=\frac{t+2}{(t+2)(t-6)}$
y=\frac{1}{(t-6)}$
$y=(t-6)^{-1}$
$y'=-(t-6)^{-2}$
$y'=\frac{-1}{(t-6)^{2}$$}
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}$$}
Find $lim_{xrightarrow infty $}$$\frac{-2x^3$$+x^2$$-2}{x^2$+10}$
Find $lim_{xrightarrow infty $}$$\frac{-2x^3$$+x^2$$-2}{x^2$+10}$
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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.
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
What is the first derivative of the function $h(x)=x^{$\frac{1}${x}$}?
What is the first derivative of the function $h(x)=x^{$\frac{1}${x}$}?
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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=alog 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]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/23829/gif.latex)
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}]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/23830/gif.latex)
frac{1}{y}cdot frac{dy}{dx}=frac{1}{x}cdot frac{1}{x} + ln xcdot frac{-1}{x^{2}}
Because we are looking for the derivative, we must solve for 
.
frac{dy}{dx}=ycdot 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 
with 
.
The answer is h'(x)=frac{x^{frac{1}{x}}}{x^{2}}(1-ln x).
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=alog x, where
ln y = frac{1}{x}ln x
We can differentiate both sides with respect to
We will need to apply the Chain Rule on the left side and the Product Rule on the right side.
frac{1}{y}cdot frac{dy}{dx}=frac{1}{x}cdot frac{1}{x} + ln xcdot frac{-1}{x^{2}}
Because we are looking for the derivative, we must solve for
frac{dy}{dx}=ycdot frac{1}{x^{2}}(1-ln x)
However, we want our answer to be in terms of
frac{dy}{dx}=frac{x^{frac{1}{x}}}{x^{2}}(1-ln x)
Since we let
The answer is h'(x)=frac{x^{frac{1}{x}}}{x^{2}}(1-ln x).