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Calculus 1 : Functions

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Example Questions

Example Question #2260 : Calculus

A very important physics formula is called the continuity equation. We will only consider the 1-dimensional continuity equation for now.

In the continuity equation, we're given that:

$\int \frac{\partial u}{\partial t} dx=Flux_{in}- Flux_{out}$ , where is a function of and .

Rewrite the right side of the equation in integral form.

Possible Answers:

$\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

$\int_{out}^{in} (Flux ) dx$

$\int_{in}^{out} (Flux ) dx$

$\int_{in}^{out} (\frac{d}{dx} Flux ) dx$

Correct answer:

$\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

Explanation:

Recall that the fundamental theorem of calculus states that:

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

Using this knowledge we write the right side as:

$Flux_{in}- Flux_{out}=$ $\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

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Example Question #1231 : Functions

Solve the following integral, where a and b are constants:

$\int abe^{b^2x}dx$

Possible Answers:

$\frac{2ae^{b^2x}}{b}+C$

$\frac{ae^{b^2x}}{b}+C$

$ae^{b^2x}+C$

$\frac{1}{2}ae^{b^2x}+C$

Correct answer:

$\frac{ae^{b^2x}}{b}+C$

Explanation:

Keeping in mind that a and b are only constants, the integral is equal to

$\int abe^{b^2x}dx=\frac{abe^{b^2x}}{b^2}+C=\frac{ae^{b^2x}}{b}+C$

and was found using the following rule:

$\int e^{ax}dx=\frac{e^{ax}}{a}+C$

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Example Question #2262 : Calculus

Evaluate the following integral:

$\int xy\cos(x^2y)dx$

Possible Answers:

$-\frac{1}{2}\sin(x^2y)+C$

$\frac{1}{2}\sin(x^2y)+C$

$\frac{1}{2}\sin(x^2y)+\frac{x^2y}{2}+C$

$\sin(x^2y)+C$

Correct answer:

$\frac{1}{2}\sin(x^2y)+C$

Explanation:

To evaluate the integral, we must make the following substitution:

u=x^2y , du=2xydx

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Rewrite the integral in terms of u and integrate:

$\frac{1}{2}\int \cos(u)du=\frac{1}{2}\sin(u)+C$

The integral was found using the following rule:

$\int \cos(x)dx=\sin(x)+C$

Finally, replace u with our original term:

$\frac{1}{2}\sin(x^2y)+C$ .

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Example Question #2263 : Calculus

In circuits with a resistor, the equation for voltage drop is given by:

$\frac{d}{dx}V =\frac{d}{dt}Q*R$ , where is voltage, is charge, and is resistance.

Write the equation as an integral expression for .

Possible Answers:

$Q=\int \frac{V}{R} dt$

$Q=\int VR dt$

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

$Q=\int {\frac{d}{dx}V}R dt$

Correct answer:

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

Explanation:

Although this may seem really difficult, we only need to solve for .

$\frac{d}{dx}V =\frac{d}{dt}Q*R$

$\frac{d}{dt}Q=\frac{\frac{\mathrm{d} }{\mathrm{d} x}V}{R}$

To solve for , integrate both sides:

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

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Example Question #1232 : Functions

$\int_{0}^{2}2x^2-5x+1 dx$

Possible Answers:

$-\frac{11}{3}$

$-\frac{8}{3}$

$-\frac{5}{3}$

$\frac{8}{3}$

Correct answer:

$-\frac{8}{3}$

Explanation:

When integrating, remember to add one to the exponent and then put that result on the denominator: $\frac{2x^3}{3}-\frac{5x^2}{2}+x$ . Now evaluate at 2, and then 0. Then subtract the two results. $(\frac{16}{3}-10+2)-0=-\frac{8}{3}$ .

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Example Question #2265 : Calculus

$\int \frac{24x^3-2x^2+4x}{4x}dx$

Possible Answers:

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

$2x^3+\frac{x^2}{4}+x+C$

$2x^3-\frac{x^2}{4}+4x+C$

$-2x^3-\frac{x^2}{4}+x+C$

$2x^3-\frac{x^2}{4}+x+C$

Correct answer:

$2x^3-\frac{x^2}{4}+x+C$

Explanation:

The first step here is to chop this up into three separate terms and then simplify since we have only one denominator: $\int 6x^2-\frac{1}{2}x+1dx$ . Then, integrate each term, remembering to add one to the exponent and then put that result on the denominator: $6(\frac{x^3}{3})-\frac{1}{2}(\frac{x^2}{2})+x$ . Simplify to get your answer: $2x^3-\frac{x^2}{4}+x+C$ . Remember to add C because it is an indefinite integral.

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Example Question #183 : Integral Expressions

$\int \frac{9+x}{x}dx$

Possible Answers:

$9ln\left | x \right |+x+C$

$9ln\left | x \right |-x+C$

$\frac{1}{9}ln\left | x \right |+x+C$

$ln\left | x \right |+x+C$

Correct answer:

$9ln\left | x \right |+x+C$

Explanation:

First, chop this expression up into two terms: $\int \frac{9}{x}+1dx$ . Then, integrate each term, remembering that when there is a single x on a denominator, the integral is $ln\left | x \right |$ . Therefore, the integration is: $9ln\left | x \right |+x+C$ . Remember to add C because it is an indefinite integral.

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Example Question #1233 : Functions

$\int_{0}^{3}2x^3-3x^2+xdx$

Possible Answers:

108

Correct answer:

Explanation:

First, integrate each term separately. Remember, when integrating, raise the exponent by one and then also put that result on the denominator: $2(\frac{x^4}{4})-3(\frac{x^3}{3})+\frac{x^2}{2}$ . Then evaluate at 3 and then 0. Subtract two results to get: $(\frac{81}{2}-27+\frac{9}{2})-0=18$ .

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Example Question #1234 : Functions

$\int \sqrt{x}dx$

Possible Answers:

$-\frac{2}{3}x^{\frac{3}{2}}+C$

$\frac{2}{3}$

$\frac{2}{3}x^{\frac{3}{2}}+C$

$\frac{2}{3}x^2+C$

Correct answer:

$\frac{2}{3}x^{\frac{3}{2}}+C$

Explanation:

To integrate this expression, I would first rewrite it to be $\int x^{\frac{1}{2}}dx$ . Then, add one to the exponent and then put that result on the denominator: $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$ . Then, simplify to get your answer of $\frac{2}{3}x^{\frac{3}{2}}+C$ . Remember to add C because it is an indefinite integral.

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Example Question #1235 : Functions

Given the one-to-one equation f(x)=3x+1, the inverse function f^-1(y)=

Possible Answers:

(y+1)/3

(y-1)/3

3x-1

undefined

(x-1)/3

Correct answer:

(y-1)/3

Explanation:

Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x."

Now we can look at the two remaining answer choices. Let y=f(x) and solve for x to find the inverse.

So f(x)=y=3x+1. Solve for x.

x=(y-1)/3

Therefore our answer is (y-1)/3.

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2260 : Calculus

Example Question #1231 : Functions

Example Question #2262 : Calculus

Example Question #2263 : Calculus

Example Question #1232 : Functions

Example Question #2265 : Calculus

Example Question #183 : Integral Expressions

Example Question #1233 : Functions

Example Question #1234 : Functions

Example Question #1235 : Functions

All Calculus 1 Resources

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