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Calculus 1 : Equations

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #151 : Writing Equations

$\int (x+1)(x^3-2)dx$

Possible Answers:

$\frac{x^5}{5}+\frac{x^4}{4}-x^2$

$-\frac{x^5}{5}-\frac{x^4}{4}-x^2+C$

$\frac{x^5}{5}+\frac{x^4}{4}-\frac{x^2}{2}+C$

$\frac{x^5}{5}+\frac{x^4}{4}-x^2+C$

Correct answer:

$\frac{x^5}{5}+\frac{x^4}{4}-x^2+C$

Explanation:

Before integrating, I would multiply the binomials (using FOIL) together to get a polynomial: $\int x^4+x^3-2x-2dx$ . Now, integrate. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator as well. You then get: $\frac{x^5}{5}+\frac{x^4}{4}-\frac{2x^2}{2}$ . Simplify and add a "C" at the end because it is an indefinite integral: $\frac{x^5}{5}+\frac{x^4}{4}-x^2+C$ .

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Example Question #152 : Writing Equations

$\int \frac{4}{x}dx$

Possible Answers:

$4ln\left | x \right |$

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

$4ln\left | x \right |+C$

$4ln\left |4 x \right |+C$

Correct answer:

$4ln\left | x \right |+C$

Explanation:

To integrate this expression, first take the 4 outside of the integral expression: $4\int \frac{1}{x}dx$ . Recall that when you are integrating $\frac{1}{x}$ , the result is $ln\left | x \right |$ . Then, multiply by 4 and tack on a "C" because it is an indefinite integral: $4ln\left | x \right |+C$ .

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Example Question #151 : Equations

$\int 2x^2-4x+1dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To integrate this expression, raise the exponent by 1 and then put that result on the denominator (while multiplying that expression by any coefficients). Therefore, when you integrate, you get: $\frac{2}{3}x^3-\frac{4x^2}{2}+x$ . Then, simplify and add "C" because it is an indefinite integral so that your answer is: $\frac{2}{3}x^3-2x^2+x+C$ .

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Example Question #154 : Integral Expressions

Evaluate the following Indefinte Integral:

$\int(Sec^2(x))-3Sin(x))dx$

Possible Answers:

Tan(x)+Cos(x)+C

Tan(x)+3Cos(x)+C

Tan^2(x)+3Cos(x)+C

-Tan(x)+3Cos(x)+C

Tan(x)-3Cos(x)+C

Correct answer:

Tan(x)+3Cos(x)+C

Explanation:

For this problem one must remember the following rule for Integrals: $\int(f(x)+g(x)))dx=\int(f(x))dx+\int (g(x))dx$

Once applying this rule we end with the following two integrals:

$\int(Sec^2(x))-3Sin(x))dx=\int (Sec^2(x))dx+\int (3Sin(x))dx$

Now apply our basic rules for Trigonometric Integrals we find the following:

Tan(x)+3Cos(x)+C

Note: a "+C" is nessecary as this integral is indefinite, that is, we are not plugging in any limits of integration.

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Example Question #155 : Integral Expressions

Evaluate the following integral:

$\int 2x^4e^{x^5}dx$

Possible Answers:

$-\frac{2}{5}e^{x^5}+C$

$2e^{x^5}+C$

$\frac{2}{5}e^{x^5}+C$

$\frac{2}{5}e^{x^5}$

Correct answer:

$\frac{2}{5}e^{x^5}+C$

Explanation:

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

u=x^5, du=5x^4dx

Next, we rewrite the integral and integrate:

$\frac{2}{5}\int e^u du=\frac{2}{5}e^u+C$

The integral was performed using the following rule:

$\int e^udu =e^u+C$

Finally, replace u with our original term:

$\frac{2}{5}e^{x^5}+C$

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Example Question #152 : Equations

Evaluate the following integral:

$\int (\sec^2(x)+x^4+3x)dx$

Possible Answers:

$-\tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C$

$\tan(x)+\frac{x^5}{5}+{3}x^2+C$

$\tan(x)+x^5+\frac{3}{2}x^2+C$

$\tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C$

Correct answer:

$\tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C$

Explanation:

The integral is equal to

$\tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C$

and was found using the following rules:

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

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Example Question #157 : Integral Expressions

Find the area under the curve $y=5x^{4}+12x^{3}+ 9x^{2}+5$ between x=1 and x=4 .

Possible Answers:

2004

1992

Correct answer:

1992

Explanation:

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

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

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{1}^{4}(5x^{4}+12x^{3}+ 9x^{2}+5 )dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= x^{5}+3x^{4}+3x^{3}+5x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(4)-F(1)

Which plugged into our integrated expression is:

$area = [(4)^{5}+3(4)^{4}+3(4)^{3}+5(4)] - [(1)^{5}+3(1)^{4}+3(1)^{3}+5(1)]$

area = [1024+768+192+20)] - [1+3+3+5]

area = [2004] - [12]

With our final answer being:

area = 1992

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Example Question #158 : Integral Expressions

Find the area under the curve $y=4x^{5}+15x^{2}+ 17x$ between x=0 and x=2 .

Possible Answers:

$-\frac{571}{6}$

$\frac{761}{6}$

222

$\frac{286}{3}$

Correct answer:

$\frac{286}{3}$

Explanation:

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

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{0}^{2}(4x^{5}+15x^{2}+ 17x)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= \frac{4}{6}x^{6}+5x^{3}+\frac{17}{2}x^{2}$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(2)-F(0)

Which plugged into our integrated expression is:

$area = [\frac{4}{6}(2)^{6}+5(2)^{3}+\frac{17}{2}+(2)^{2}] - [\frac{4}{6}(0)^{6}+5(0)^{3}+\frac{17}{2}(0)^{2}]$

$area = [\frac{4}{6}(64)+5(8)+\frac{17}{2}+(4)] - [\frac{4}{6}(0)+5(0)+\frac{17}{2}(0)]$

$area = [\frac{256}{6}+40+\frac{17}{2}+4] - [0+0+0]$

$area = [\frac{256}{6}+\frac{240}{6}+\frac{51}{6}+\frac{24}{6}] - [0]$

$area = [\frac{571}{6}] - [0]$

With our final answer being:

$area = \frac{571}{6}=\frac{286}{3}$

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Example Question #159 : Integral Expressions

Find the area under the curve y=15x between x=3 and x=5 .

Possible Answers:

120

$\frac{375}{2}$

Correct answer:

120

Explanation:

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

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{3}^{5}(15x)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= \frac{15}{2}x^{2}$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(5)-F(3)

Which plugged into our integrated expression is:

$area = [\frac{15}{2}(5)^{2}] - [\frac{15}{2}(3)^{2}]$

$area = [\frac{15}{2}(25)] - [\frac{15}{2}(9)]$

$area = [\frac{375}{2}] - [\frac{135}{2}]$

$area = \frac{240}{2}$

With our final answer being:

area = 120

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Example Question #160 : Integral Expressions

Find the area under the curve $y=32x^{3}+18x^{2}+3x+42$ between x=-1 and x=1 .

Possible Answers:

115

Correct answer:

Explanation:

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

We plug in the equation of the curve in for f(x) , the smaller of our x-values for , and the larger of our x-values for . For this problem, our set-up looks like this:

$\int_{-1}^{1}(32x^{3}+18x^{2}+3x+42)dx$

Next, we integrate our expression, ignoring our and values for now. We get:

$F(x)= 8x^{4}+6x^{3}+\frac{3}{2}x^{2}+42x$

Now, to find the area under the part of the curve we're looking for, we plug our and values into our integrated expression and find the difference, using the following skeleton:

area = F(b)-F(a)

In this problem, this looks like:

area = F(1)-F(-1)

Which plugged into our integrated expression is:

$area = [8(1)^{4}+6(1)^{3}+\frac{3}{2}(1)^{2}+42(1)] - [8(-1)^{4}+6(-1)^{3}+\frac{3}{2}(-1)^{2}+42(-1)]$

$area = [8+6+\frac{3}{2}+42] - [8(1)+6(-1)+\frac{3}{2}(1)+42(-1)]$

$area = [\frac{16}{2}+\frac{12}{2}+\frac{3}{2}+\frac{84}{2}] - [\frac{16}{2}-\frac{12}{2}+\frac{3}{2}-\frac{84}{2}]$

$area = [\frac{115}{2}] - [-\frac{77}{2}]$

$area = \frac{192}{2}$

With our final answer being:

area = 96

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #151 : Writing Equations

Example Question #152 : Writing Equations

Example Question #151 : Equations

Example Question #154 : Integral Expressions

Example Question #155 : Integral Expressions

Example Question #152 : Equations

Example Question #157 : Integral Expressions

Example Question #158 : Integral Expressions

Example Question #159 : Integral Expressions

Example Question #160 : Integral Expressions

All Calculus 1 Resources

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