Fundamental Theorem of Calculus - AP Calculus AB

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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.

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Question

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

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?

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:

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

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 ?

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 .

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 ?

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?

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

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

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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

.

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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 $\frac{1}{4}$ .

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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 ?

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 .

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 ?

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?

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

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

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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Question

Using the Fundamental Theorem of Calculus solve the integral.

$\int_{1}^{3}e^xdx$

Answer

To solve the integral using the Fundamental Theorem, we must first take the anti-derivative of the function. The anti-derivative of is . Since the limits of integration are 1 and 3, we must evaluate the anti-derivative at these two values.

denotes the anti-derivative.

When we do this,

and .

The next step is to find the difference between the values at each limit of integration, because the Fundamental Theorem states

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

Thus, we subtract to get a final answer of .

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Question

Using the Fundamental Theorem of Calculus and simplify completely solve the integral.

$\int_{3}^{6}\frac{dx}{x}$

Answer

To solve the integral, we first have to know that the fundamental theorem of calculus is

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

Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6.

To find the anti-derivative, we have to know that in the integral, $\frac{dx}{x}$ is the same as $\frac{1}{x}dx$ .

The anti-derivative of the function $\frac{1}{x}$ is , so we must evaluate .

According to rules of logarithms, when subtracting two logs is the same as taking the log of a fraction of those two values:

$ln\frac{6}{3}$ .

Then, we can simplify to a final answer of

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Question

Solve $\int_{0}^{3}(x^3-6x)dx$ using the Fundamental Theorem of Calculus.

Answer

To solve the integral, we first have to know that the fundamental theorem of calculus is

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

Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 0 and 3.

The anti-derivative of the function is $\frac{1}{4}x^4-3x^2$ , so we must evaluate .

When we plug 3 into the anti-derivative, the solution is $\frac{-27}{4}$ , and when we plug 0 into the anti-derivative, the solution is 0.

To find the final answer, we must take the difference of these two solutions, so the final answer is $\frac{-27}{4}$ .

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