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

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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Question

Solve $\int_{0}^{2}(2x^3-6x+\frac{3}{x^2+1})$ 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 2.

The anti-derivative of the function

$2x^3-6x+\frac{3}{x^2+1}$

is

$\frac{1}{2}x^4-3x^2+3tan^{-1}x$ ,

so we must evaluate .

When we plug 3 into the anti-derivative, the solution is $-4+3tan^{-1}(2)$ , 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 $-4+3tan^{-1}(2)$ .

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Question

Use the fundamental theorem of Calculus to evaluate the definite integral

$\int_{\pi/2}^{\pi}(2cos(x)+1)dx$

Answer

Here we use the fundamental theorem of Calculus: $\int_{a}^{b}f(x)dx=F(b)-F(a)$

$F(x)=\int(2cos(x)+1)dx=2sin(x)+x$

Here we do not worry about adding a constant c because we are evaluating a definite integral.

$F(b)=F(\pi)=2sin(\pi)+\pi=0+\pi=\pi$

$F(a)=F(\frac{\pi}{2})=2sin(\frac{\pi}{2})+\frac{\pi}{2}= 2+\frac{\pi}{2}$

$F(b)-F(a)=F(\pi)-F(\frac{\pi}{2})=\pi-(2+\frac{\pi}{2})=\frac{\pi}{2}-2$

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Question

Evaluate $\int_1^x\frac{2}{t^2}dt$ .

Answer

We can integrate this without too much trouble

$\int_1^x\frac{2}{t^2}dt$ . Start

$=\int_1^x2t^{-2}dt$ . Rewrite the power

$=[-2t^{-1}]_1^x$ . Integrate

$=(-2x^{-1})- (-2)$ . Evaluate

$=\frac{2(x-1)}{2}$ . Simplify

Note that we were not asked to evaluate $\frac{d}{dx}\int_1^x\frac{2}{t^2}dt$ , so you should not attempt to use part one of the Fundamental Theorem of Calculus. This would give us the incorrect answer of $\frac{2}{x^2}$ .

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Question

Evaluate the indefinite integral:

$\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)dx$

Answer

$\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)dx$

First compute the indefinite integral:

$\int \tan(x)\sec^2(x)dx$

Note that the $\sec^2(x)$ is the derivative of $\tan(x)$ . So proceed by defining a new variable:

$u = \tan(x) : : : : :\rightarrow : : : : :du=\sec^2(x)dx$

Now the the integral can be written in terms of

$\int \tan(x)\sec^2(x)dx=\int udu =\frac{1}{2}u^2 +C=\frac{1}{2}\tan^2(x)+C$

Therefore:

$\int \tan(x)\sec^2(x)dx=\frac{1}{2}\tan^2(x)+C$

When we go to compute the indefinite integral the constant of integration will be ignored since it will be subtracted out when we evaluate.

We can precede by either going back to the original variable and evaluate over the original limits of integration, or we can find new limits of integration corresponding to the new variable . Let's look at both equivalent methods:

Solution 1)

$\int_0^{\frac{\pi}{4}}\tan(x)\sec^2(x)dx=\left[ \frac{1}{2}\tan^2(x)\right ]_{x=0}^{x=\frac{\pi}{4}}: : =: : : : \frac{1}{2}\tan^2\left(\frac{\pi}{4}\right) -\frac{1}{2}\tan^2\left(0\right)$

$\tan(0)=0$ so the last term vanishes. The first term reduces to $\frac{1}{2}$ since the tangent function is equal to .

$\int_o^{\frac{\pi}{4}}\tan(x)\sec^2(x)dx=\frac{1}{2}$

Solution 2)

We could have also solved without converting back to the original variable. Instead, we could just change the limits of integration. Use the definition assigned to the variable , which was $u = \tan(x)$ and then use this to find which value takes on when (lower limit) and when $x =\frac{\pi}{4}$ (upper limit).

$x = 0: : : : : : :\rightarrow: : : : :u = \tan(0)=0$

$x=\frac{\pi}{4}: : : : :\rightarrow : : : : :u=tan\left(\frac{\pi}{4}\right )=1$

$\int_0^1 udu =\left[\frac{1}{2}u^2 \right ]_{u=0}^{u=1} =\frac{1}{2}$

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Question

$\frac{d}{dx}\int_{0}^{x^2} sin(t)dt=?$

Answer

Derivatives and anti derivatives annihilate each other. Therefore, the derivative of an anti derivative is simply the function in the integral with the limits substituted in multiplied by the derivative of each limit. Also, be careful that the units will match the outermost units (which comes from the derivative).

$\frac{d}{dx}\int_{0}^{x^2} sin(t)dt= sin(x^2)\frac{d}{dx}(x^{2})-sin(0)\frac{d}{dx}(0)'$

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Question

$\frac{d}{dx}\int_{x}^{x^3}\frac{1}{t^2+1}dt=?$

Answer

This is a Fundamental Theorem of Calculus problem. Since a derivative and anti-derivative cancel each other out, we simply have to plug the limits into our function (with the outside variable). Then, we multiply each by the derivative of the bound:

$\frac{d}{dx}\int_{x}^{x^3}\frac{1}{t^2+1}dt=\frac{1}{{(x^3)}^2+1}\frac{d}{dx}(x^3)-\frac{1}{{(x)}^2+1}\frac{d}{dx}(x)$

$=\frac{3x^2}{x^6+1}-\frac{1}{x^2+1}.$

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Question

$\frac{d}{dx}\int_{0}^{sinx}(1-t^2)dt=?$

Answer

Using the Fundamental Theorem of Calculus, the derivative of an anti-derivative simply gives us the function with the limits plugged in multiplied by the derivative of the respective bounds:

$\frac{d}{dx}\int_{0}^{sinx}(1-t^2)dt=(1-sin^2x)\frac{d}{dx}(sinx)-(1-0^2)\frac{d}{dx}(0)$

In the last step, we made use of the following trigonometric identity:

$sin^2x+cos^2x=1. \rightarrow1-sin^2x=cos^2x.$

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Question

Evaluate the following indefinite integral:

$\int4t^3-6sin(t) dt$

Answer

Evaluate the following indefinite integral:

$\int4t^3-6sin(t) dt$

Recall that we can split subtraction and addition within integrals into separate integrals. This means that we can look at our problem in two steps.

Recall that we can integrate any exponential term by adding 1 to the exponent and dividing by the new exponent.

So,

$\int 4t^3dt=\frac{4t^{3+1}}{4}=t^4+c$

Next, recall that the integral of sine is negative cosine. However, we already have a negative sine, so we should get positive cosine.

$\int -6sin(t)dt=6cos(t)+c$

Now, we can combine our two halves to get our final answer.

Notice that we only have one "c" because c is just a constant, not a variable.

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Question

Evaluate the given definite integral:

$F(x)=\int^5_0 5e^x-5x^4+cos(x)dx$

Answer

Evaluate the given definite integral:

$\int^5_0 5e^x-5x^4+cos(x)dx$

Let's look at this integral as three separate parts, then we'll combine those parts to evaluate the answer.

First, the integral of is just

Second, The integral of any exponential term can be found by increasing the exponent by 1 and then dividing by the new number.

Third, the integral of cosine is simply positive sine.

Put all this together to get:

Now, we need to evaluate our answer by finding the difference between the limits of integration.

$5e^x-x^5+sin(x)+c\mid^5_0$

First, let's see what we get when we plug in 0

Next, let's use 5

Next, find the difference of F(5)-F(0)

So our answer is: -2388.89.

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