Fundamental Theorem of Calculus - AP Calculus AB

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Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

← Didn't Know|Knew It →

Question

Evaluate the integral $\int_{0}^{\pi/2}\cos(x)-2\sin(2x)dx$

Tap to reveal answer

Answer

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

$\int_a^b f(x)dx = F(b)-F(a)$ , where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

$\int_0^{\pi/2}\cos(x)dx - \int_0^{\pi/2}2\sin(2x)dx$

First, the integral of the cosine is the sine. Doing this, we get

$[\sin(x)]|_0^{\pi/2} - \int_0^{\pi/2}2\sin(2x)dx$

The $|_0^{\pi/2}$ is just notation for "evaluated from to $\pi/2$ ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The $[\sin(x)]$ is the from the theorem. Before moving to the second integral, we can apply this theorem.

$F(\pi/2)-F(0)$

$\sin(\pi/2)- \sin(0)$

So far, we have the following expression for the entire problem

$1- \int_0^{\pi/2}2\sin(2x)dx$

The basic integral form for the remaining integral is $\int_a^b \sin(u)du =[ -\cos(u)]|_0^{\pi/2}$

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. $1 - \int_0^{\pi/2}2\sin(2x)dx$ $1- [-\cos(2x)]|_0^{\pi/2}$

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in $\pi/2$ . Plug in and subtract.

$1- \{[-\cos(2\cdot \pi/2)]-[-\cos(2\cdot0)]\}$

$1-\{[-\cos(\pi)]-[-\cos(0)]\}$

$1- \{[-(-1)]-[-(1)]\}$

$1-\{1+1\}$

This is our answer.

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Question

Write the domain of the function.

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

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Answer

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

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

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Question

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

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Answer

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

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Question

Which of the following is a vertical asymptote?

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Answer

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

Vertical asymptotes occur at values. The horizontal asymptote occurs at

.

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Question

Evaluate the following limit:

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

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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 $\frac{1}{4}$ .

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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 ?

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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 $-\frac{1}{2}$ .

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Question

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

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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 $\int x^n\ dx =\frac{x^{n+1}}{n+1}+C$ .

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

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

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

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Question

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

What are the horizontal asymptotes of ?

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Answer

Compute the limits of as approaches infinity.

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Question

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

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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 $\frac{63}{4}$ .

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Question

Evaluate

$\int_{-1}^{e-2}\frac{1}{x+2}$

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Answer

We will use the Fundamental Theorem of Calculus

$\int_{a}^{b}f(x)dx=F(b)-F(a)$

and the rule

$\int\frac{1}{x+a}dx=\ln(x+a)$

First we find the anti derivative

$\int_{-1}^{e-2}\frac{1}{x+2}=\ln(x+2)|_{-1}^{e-2}$

And then we evaluate, (upper minus lower)

$\ln(x+2)|_{-1}^{e-2}=\ln(e-2+2)-\ln(-1+2)=\ln(e)-\ln(1)$

(Remembering your logarithm rules)

$\ln(e)-\ln(1)=1-0=1$

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