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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Computation Of Derivatives

Find the derivative of the function $y = \int_{0}^{x^2} e^{-t^2}dt$

Possible Answers:

None of the other answers.

x^2

$2xe^{-x^4}$

$e^{-x^2}$

Correct answer:

$2xe^{-x^4}$

Explanation:

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$y = \int_{0}^{x^2} e^{-t^2}dt$ . Start

$y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$ . Take the derivative of both sides with respect to .

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's in this case), and that the upper limit of the integral is a function of , (it's x^2 in this case).

Afterward, plug x^2 in for , and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$ .

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Example Question #5 : Other Derivative Review

Find the derivative of $y = \int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ .

Possible Answers:

Does not exist

None of the other answers

$\frac{1}{\sqrt{1-x^2}}$

$\sin^{-1}(\sqrt{x})$

Correct answer:

$\sin^{-1}(\sqrt{x})$

Explanation:

Use the Fundemental Theorem of Calculus to find .

$y = \int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ . Start

$y' = \frac{d}{dx}\int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ . Take derivatives of both sides with respect to .

Use the Fundemental Theorem of Calculus

$\frac{d}{dx}\int^{f(x)}_{x_0} g(t) dt = g(f(x))\times f'(x)$

with $f(x) = x, g(t) = \sin^{-1}(\sqrt{t})$ , x_0 = 0 to obtain

$y' = \sin^{-1}(\sqrt{x})$ .

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Example Question #2 : Computation Of Derivatives

Find the derivative of the function $y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$

Possible Answers:

Does not exist

None of the other answers

$\frac{2xe^x\cos{x}}{x}$

$x^2e^x\ln(x)\sin(x)$

Correct answer:

$x^2e^x\ln(x)\sin(x)$

Explanation:

To find the derivative of this function, we need to use the Fundemental Theorem of Calculus Part 1 (As opposed to the 2nd part, which is what's usually used to evaluate definite integrals)

$y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Start

$y'= \frac{d}{dx}\int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Take derivatives of both sides.

$y'= x^2e^x\ln(x)\sin(x)$ . "Cancel" the integral and the derivative. (Make sure that the upper bound on the integral is a function of , and that the lower bound is a constant before you cancel, otherwise you may need to use some manipulation of the bounds to make it so.)

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Example Question #11 : Other Derivative Review

Let f(x)=cos^3(x+1) . Determine the value of $f'(\pi)$ .

Possible Answers:

$-3cos^2(\pi+1)sin(\pi+1)$

$3cos^2(\pi+1)sin(\pi+1)$

$-3cos^2(\pi+1)$

$3cos^2(\pi+1)$

Correct answer:

$-3cos^2(\pi+1)sin(\pi+1)$

Explanation:

Let us first determine the derivative of f(x) by successive applications of the chain rule. The given function is a composition of three simpler functions: first add to , then take the cosine of the sum, and then cube the result. By the chain rule, we must differentiate with respect to each of these functions and take their product to obtain the derivative of the original function, as shown below:

f(x)=cos^3(x+1),

f'(x)=(3cos^2(x+1))(-sin(x+1))(1)

= -3cos^2(x+1)sin(x+1)

To determine the value of this function at $x=\pi$ , simply substitute $\pi$ for :

$-3cos^2(\pi+1)sin(\pi+1)$

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Example Question #12 : Other Derivative Review

Evaluate.

$\int \ \cos(2x ) \ dx$

Possible Answers:

$F(x) = \frac{\sin (2x)}{2}$

$F(x) = \frac{\cos (2x)}{2}$

$F(x) = - \frac{\sin (x)}{2}$

Answer not listed

$F(x) = - \frac{\cos (x)}{2}$

Correct answer:

$F(x) = \frac{\sin (2x)}{2}$

Explanation:

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

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \cos(2x)$ .

The antiderivative is $F(x) = \frac{\sin (2x)}{2}$ .

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Example Question #1611 : Calculus Ii

Evaluate.

$\int \sqrt{x-2} \ dx$

Possible Answers:

$F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$

$F(x) = - \frac{(x-2)^{\frac{3}{2}}}{3}$

$F(x) = (x-2)^{\frac{3}{2}}$

Answer not listed

$F(x) = (x-2)^{\frac{1}{2} }$

Correct answer:

$F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$

Explanation:

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

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \sqrt{x-2}$ and u = x-2 .

The antiderivative is $F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$ .

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Example Question #1612 : Calculus Ii

Evaluate.

$\int e^{3x-1} \ dx$

Possible Answers:

$F(x) = e^{3x-1}$

Answer not listed

$F(x) = \frac{e^{x}}{3x-1}$

$F(x) = \frac{e^{3x-1}}{3}$

$F(x) =3 e^{3x-1}$

Correct answer:

$F(x) = \frac{e^{3x-1}}{3}$

Explanation:

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

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = e^{3x-1}$ and u = 3x-1 .

The antiderivative is $F(x) = \frac{e^{3x-1}}{3}$ .

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Example Question #15 : Other Derivative Review

Suppose and are related implicitly by the equation x^3+y+y^3=x . Find $\frac{d^2y}{dx^2}$ in terms of and .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To take the second derivative of this implicit function, we must take the first derivative. To do so, we take the derivative of our implicit function with respect to :

$\frac{d}{dx}(x^3+y+y^3)=\frac{d}{dx}(x)$

$3x^2+\frac{dy}{dx}+3y^2\frac{dy}{dx}=1$

Here, we invoked the chain rule to take the derivatives of the terms, imagining as the inner function. This gives us an equation in $\frac{dy}{dx}$ which we can then solve for $\frac{dy}{dx}$ by algebraic means:

$\frac{dy}{dx}+3y^2\frac{dy}{dx}=1-3x^2$

$\frac{dy}{dx}(1+3y^2)=1-3x^2$

$\frac{dy}{dx}=\frac{1-3x^2}{1+3y^2}$

Now we have the value of $\frac{dy}{dx}$ . Now to find $\frac{d^2y}{dx^2}$ , we take the derivative of the above function with respect to :

$\frac{d^2x}{dy^2}=\frac{d}{dx}\biggr(\frac{1-3x^2}{1+3y^2}\biggr)$

$\frac{d^2x}{dy^2}=\frac{\big[\frac{d}{dx}(1-3x^2)\big](1+3y^2)-(1-3x^2)\big[\frac{d}{dx}(1+3y^2)\big]}{(1+3y^2)^2}$

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\big(6y\frac{dy}{dx}\big)}{(1+3y^2)^2}$

In the above, we used the quotient rule to take the derivative of the fraction, and then again using the chain rule on expressions involving . Now since our derivative must be in terms of and , we need to get rid of the $\frac{dy}{dx}$ in the above equation. Thus we substitute the first derivative expression $\frac{1-3x^2}{1+3y^2}$ which we found at the start of the problem into the fraction above:

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\bigg(6y\big(\frac{1-3x^2}{1+3y^2}\big)\bigg)}{(1+3y^2)^2}$

Now it is a matter of simplifying these equations. To do so, we multiply top and bottom by 1+3y^2 :

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\bigg(6y\big(\frac{1-3x^2}{1+3y^2}\big)\bigg)}{(1+3y^2)^2}\frac{1+3y^2}{1+3y^2}$

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)^2-(1-3x^2)6y(1-3x^2)}{(1+3y^2)^3}$

$\frac{d^2x}{dy^2}=-\frac{6x(1+3y^2)^2+6y(1-3x^2)^2}{(1+3y^2)^3}$

Finally, we arrive at the desired result.

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Example Question #1612 : Calculus Ii

Evaluate.

$\int x^4 \ dx$

Possible Answers:

$F(x) = \frac{1}{3}x^3$

Answer not listed

$F(x) = \frac{1}{4}x^4$

F(x) = x^4

$F(x) = \frac{1}{4}x$

Correct answer:

$F(x) = \frac{1}{4}x^4$

Explanation:

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

In order to evaluate this integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = x^4 .

The antiderivative is $F(x) = \frac{1}{4}x^4$ .

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Example Question #491 : Derivatives

What is the derivative of tan(x) ?

Possible Answers:

$\frac{-1}{\sqrt{1-x^2}}+C$

sec^2(x)

sec(x)

cot(x)

$\frac{1}{\sqrt{1-x^2}}$

Correct answer:

sec^2(x)

Explanation:

The trigonometric functions have specific derivatives that one needs to memorize.

The basic trigonometric derivatives are as follows.

$\\ sin(x)\frac{d}{dx}=cos(x) \\ cos(x)\frac{d}{dx}=-sin(x) \\ tan(x)\frac{d}{dx}=sec^2(x)$ $\\csc(x)\frac{d}{dx}=-csc(x)cot(x) \\ \\sec(x)\frac{d}{dx}=sec(x)tan(x) \\ \\cot(x)=-csc^2x$

This particular question asks for the derivative for tangent and thus the correct solution is,

$tan(x)\frac{d}{dx}=sec^2(x)$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Computation Of Derivatives

Example Question #5 : Other Derivative Review

Example Question #2 : Computation Of Derivatives

Example Question #11 : Other Derivative Review

Example Question #12 : Other Derivative Review

Example Question #1611 : Calculus Ii

Example Question #1612 : Calculus Ii

Example Question #15 : Other Derivative Review

Example Question #1612 : Calculus Ii

Example Question #491 : Derivatives

All Calculus 2 Resources

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