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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #7 : Area

Find the value of $\int_{0}^{\infty}xe^{-x^2}dx.$

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To solve this problem, we will need to do a -substitution. Letting

$u=-x^2\Rightarrow \frac{du}{dx}=-2x\Rightarrow \frac{-du}{2}=xdx$ .

Substituting our u(x) function back into the integral, we get

$\int \frac{-1}{2}e^udu=\frac{-1}{2}e^u=\frac{-1}{2}e^{-x^2}.$

Evaluating this at $x=\infty$ and x=0 we get

$\frac{-1}{2}e^{- \infty^2}-(\frac{-1}{2}e^0)=0-(\frac{-1}{2})=\frac{1}{2}$

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Example Question #8 : Area

Find the average value of x^3 - 2x^2 +x on the interval (0, 1).

Possible Answers:

1/6

-1/3

1/3

1/12

Correct answer:

1/12

Explanation:

The average is given by integration as:

$\text{avg}_{A, B}[ f ] = \frac{1}{B - A} \int_A^B f(x)~dx$

This means that:

$\text{avg}_{0, 1}[x \to x^3 - 2x^2 + x ] = \frac{1}{1 - 0} \int_0^1 [x^3 - 2x^2 + x]~dx$

$= \left[\frac{x^4}{4} - \frac{2 x^3}{3} + \frac{x^2}{2}\right]_0^1 = \frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{1}{12}$

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Example Question #4001 : Calculus

We have the function $f(x)=\sqrt{x}$ and it is used to form a three-dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Possible Answers:

66.667

78.832

56.111

Area is infinite

Correct answer:

78.832

Explanation:

Imagine a point somewhere on the function $f(x)=\sqrt{x}$ and it rotates about $y=4$ to form a circle with a circumference of $2\pi (4-\sqrt{x})$ where $(4-\sqrt{x})$ is the radius. You may have to draw a picture/graph to make sure this is clear.

Next, pretend that this circle is a circular strip with thickness $\Delta x$ . To find the area of that circular strip, we times that thickness by the circumference so that we have

$2\pi (4-\sqrt{x})\Delta x$

Now, imagine that the three dimensional figure is made up of many of these circular strips. To find the total volume, we need to sum up the areas of all of these strips. We do this by turning $2\pi (4-\sqrt{x})\Delta x$ into an integral from 0 to 5.

$Volume=2\pi \int_{0}^{5}(4-\sqrt{x})dx$

Perform your integration

$Volume=2\pi \left [ 4(5)-\frac{2}{3}(5)^{\frac{3}{2}} - 4(0)+\frac{2}{3}(0)^{\frac{3}{2}} \right ]=2\pi\left [ 20-\frac{2}{3}\sqrt{5^3} \right ]$

Volume=78.832

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Example Question #4002 : Calculus

Calculate the area between the parabola f(x)=4+5x-x^2 and the line f(x)=x-1 .

Possible Answers:

$\frac{100}{3}$

$\frac{98}{3}$

Correct answer:

Explanation:

To calculate the area between these two functions, we are going to need to set up a definite integral, so our first step is to see where our two boundaries intersect. We do this by setting our two functions equal to each other, and solving for the x values at which they intersect:

4+5x-x^2=x-1

x^2-4x-5=(x-5)(x+1)=0

x=5,x=-1

Here we can see that our two functions intersect at x=5 and x= -1, so these will be the limits for our definite integral. Now that we know our limits, we set up our integral of the function of the upper boundary minus the function for the lower boundary:

$\int_{-1}^{5}((4+5x-x^2)-(x-1))dx=\int_{-1}^{5}(-x^2+4x+5)dx$

$=(-\frac{1}{3}x^3+2x^2+5x)_{-1}^{5}$

$=(-\frac{1}{3}(5)^3+2(5)^2+5(5))-(-\frac{1}{3}(-1)^3+2(-1)^2+5(-1))$

$=(-\frac{125}{3}+50+25)-(\frac{1}{3}+2-5)=(\frac{100}{3})-(-\frac{8}{3})=36$

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Example Question #12 : Area

Find the area under the curve of y=1 from [-1,1] .

Possible Answers:

Correct answer:

Explanation:

To find the area under the curve, integrate the function and evaluate at the bounds.

$\int_{-1}^{1} 1 dx= x|_{-1}^{1} = 1-(-1) = 2$

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Example Question #14 : How To Find Area Of A Region

Find the area bounded by y=x and $y=\sqrt x$ .

Possible Answers:

$\frac{1}{4}$

$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{6}$

Explanation:

The top curve will be $y=\sqrt x$ , and the bottom curve will be y=x . The area is the integral of the top minus the bottom curve.

First, determine the bounds of integration by setting both equations equal to each other, and solve for x. These values are where both graphs intersect.

$x= \sqrt x$

x^2 = x

x^2-x=0

x(x-1)=0

x=0

x=1

The bounds of integration will be from 0 to 1. Setup and evaluate the integral.

$\int_{0}^{1}\sqrt(x)-x\:dx= \left[\frac{2}{3}x^\frac{3}{2} -\frac{1}{2}x^2\right]_{0}^{1} = \frac{2}{3}-\frac{1}{2}=\frac{1}{6}$

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Example Question #4007 : Calculus

What is the area under the curve bounded by y=x^3 from [1,3] ?

Possible Answers:

$\frac{81}{4}$

Correct answer:

Explanation:

The area under the curve is the integral of the function evaluated at the interval given.

Write the integral to be evaluated.

$\int_{1}^{3}x^3\:dx = \left[\frac{1}{4}x^4\right]_{1}^{3} = \frac{1}{4}(3)^4-\frac{1}{4}(1)^4$

$\frac{81}{4}-\frac{1}{4} =\frac{80}{4}=20$

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Example Question #4008 : Calculus

Find the area of the region bounded by the following two curves:

y=x^2

$y=\sqrt{x}$

Possible Answers:

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

In order to find the area of the region bounded by the two curves, we must first find the bounds of the region by determning where the curves intersect:

$x^2=\sqrt{x}\rightarrow x^2-\sqrt{x}=0\rightarrow x^{\frac{1}{2}}(x^{\frac{3}{2}}-1)=0$

$x^{\frac{1}{2}}=0\rightarrow x=0$

$x^{\frac{3}{2}}-1=0\rightarrow x^{\frac{3}{2}}=1\rightarrow x=1$

The curves intersect at x=0 and x=1, so these are the bounds of the region for which we want to determine the area. Our next step is to set up an integral with these bounds, where the bottom curve is subtracted from the upper curve, which can then be evaluated to find the area of the region:

$A=\int_{0}^{1}(\sqrt{x}-x^2)dx=\left[\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^3\right]_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

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Example Question #4009 : Calculus

Find the area of the region bounded by the following two functions:

y=4x+16

y=2x^2+10

Possible Answers:

$\frac{64}{3}$

$\frac{17}{2}$

$\frac{54}{5}$

Correct answer:

$\frac{64}{3}$

Explanation:

Before we can find the area of the region bounded by the two curves, we must first determine the bounds of that region by finding where the two functions intersect. We do this by setting them equal to each other and solving for x:

$4x+16=2x^2+10\rightarrow 2x^2-4x-6=0\rightarrow 2(x^2-2x-3)=0$

$2(x-3)(x+1)=0\rightarrow x=-1,x=3$

The functions intersect at x=-1 and x=3, so these are the bounds of the area. Now we set up an integral with these bounds, where the lower function is subtracted from the upper function:

$A=\int_{-1}^{3}[(4x+16)-(2x^2+10)]dx=\int_{-1}^{3}(4x-2x^2+6)dx$

$=\left[2x^2-\frac{2}{3}x^3+6x\right]_{-1}^{3}=(18-18+18)-\left(2+\frac{2}{3}-6\right)=\frac{64}{3}$

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Example Question #98 : Regions

In units squared, what is the area underneath f(x) on the interval $\left \{ 0,5\right \}$ ?

f(x)=x^2+7

Possible Answers:

$76\frac{3}{4}$

$76\frac{1}{3}$

$76\frac{5}{11}$

$76\frac{2}{3}$

Correct answer:

$76\frac{2}{3}$

Explanation:

To find the area under f(x), we need to evaluate the following:

$\int_{0}^{5}x^2+7dx={\frac{x^3}{3}+7x+c\mid }\binom{5}{0}$

To Evaluate, perform the following:

$(\frac{5^3}{3}+(7*5)+c)-(\frac{0^3}{3}+(7*0)+c)=\frac{5^3}{3}+35=76\frac{2}{3}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #7 : Area

Example Question #8 : Area

Example Question #4001 : Calculus

Example Question #4002 : Calculus

Example Question #12 : Area

Example Question #14 : How To Find Area Of A Region

Example Question #4007 : Calculus

Example Question #4008 : Calculus

Example Question #4009 : Calculus

Example Question #98 : Regions

All Calculus 1 Resources

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