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Calculus 1 : Calculus

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

Example Questions

Example Question #176 : Regions

Find the area of the region bounded by the curve $\small f(x)=9x^2+14x+3$ and the -axis over the interval $\small x=[1,3]$ .

Possible Answers:

$\small 83$

$\small 140$

$\small 13$

$\small 153$

$\small 70$

Correct answer:

$\small 140$

Explanation:

The area of the region bounded underneath by the x-axis can be found by integrating the function

$\small f(x)=9x^2+14x+3$

over the specified interval of $\small x=[1,3]$ :

To integrate use the rule,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule to our function we find the following.

$\small A=\int_1^3 (9x^2+14x+3)dx$

$\small A=3x^3+7x^2+3x|_1^3$

$\small A=(81+63+9)-(3+7+3)$

$\small A=153-13$

$\small A=140$

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Example Question #177 : Regions

Find the area of the region bounded by the functions $\small \small f(x)=6cos(\frac{\pi}{6}x)$ , $\small \small g(x)=\frac{3}{2}x$ , and on the left by the -axis.

Possible Answers:

$\small \frac{18\sqrt3-3\pi}{\pi}$

$\small \frac{18\sqrt3+3\pi}{\pi}$

$\small 18\pi\sqrt3+3$

$\small 18\sqrt3-3\pi$

$\small 18\pi\sqrt3-3\pi$

Correct answer:

$\small \frac{18\sqrt3-3\pi}{\pi}$

Explanation:

To find the area of the region between the functions $\small \small f(x)=6cos(\frac{\pi}{6}x)$ , $\small \small g(x)=\frac{3}{2}x$ , the first step will be to determine the lower and upper x bounds. We're told the region is bounded by the y-axis on the left, so $\small x_l=0$ . The upper x-bound will be where the two functions intersect:

$\small 6cos(\frac{\pi}{6}x_u)=\frac{3}{2}x_u$

For the value of $\small x_u=2$ , each of the functions equals $\small 3$ .

Over the interval of $\small x=[0,2]$ , $\small f(x)>g(x)$ , so the area of the region can be determined by the integral:

Use the following rules to integrate the function:

$\int x^n=\frac{x^{n+1}}{n+1}$ and $\int cos(u)dx=\frac{1}{\frac{du}{dx}}sin(u)$

Applying the above rules we get,

$\small A=\int_0^2 (6cos(\frac{\pi}{6}x)-\frac{3}{2}x)dx$

$\small A=\frac{36}{\pi}sin(\frac{\pi}{6}x)-\frac{3}{4}x^2|_0^2$

$\small A=(\frac{36}{\pi}(\frac{\sqrt3}{2})-3)-(0-0)$

$\small A=\frac{18\sqrt3-3\pi}{\pi}$ .

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Example Question #101 : Area

Find the area of the region bounded by the function $\small f(x)=4-4e^{-2x}$ and the -axis over the interval $\small x=[0,2]$ .

Possible Answers:

$\small 8+2e^{-4}$

$\small 10-2e^{-4}$

$\small 10+2e^{-4}$

$\small 8-2e^{-4}$

$\small 6+2e^{-4}$

Correct answer:

$\small 6+2e^{-4}$

Explanation:

To find the area of the region between $\small f(x)=4-4e^{-2x}$ and the x-axis, integrate the function over the specified interval:

For this particular function use the following rules to integrate.

$\int x^n=\frac{x^{n+1}}{n+1}$ and $\int e^u=\frac{1}{\frac{du}{dx}}e^u$

Therefore we get,

$\small A=\int_0^2(4-4e^{-2x})dx$

$\small A=4x+2e^{-2x}|_0^2$

$\small A=(8+2e^{-4})-(4(0)+2e^{-2(0)})$

$\small A=8+2e^{-4}-2$

$\small \small A=6+2e^{-4}$ .

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Example Question #102 : How To Find Area Of A Region

A construction company is pouring concrete pads for an abstract art-themed promenade on a college campus. One slab's overhead view is given by region contained between the functions f(x)=x+2 and $g(x)=x^{2}-4$ and to the right of the -axis, where gives the size in feet. If the slab is two feet deep, what is the volume occupied by the concrete slab?

Possible Answers:

$\frac{68}{3}ft^{3}$

$\frac{27}{2}ft^{3}$

$54 ft^{3}$

$27 ft^{3}$

Correct answer:

$27 ft^{3}$

Explanation:

Since the depth of the slab is constant, we know it's volume is given by:

V= Ah =2A

The area is, of course given by an integral. To calculate it, we must know where the two functions intersect:

$x+2 = (x^{2}-4)$

$x^{2}-x-6 = 0$

Which has roots x=-2 and x=3 . Since we are concerned with area to the right of the y-axis, we discard the negative root. So the area is given by:

$A=\int_{0}^{3}x+2-(x^{2}-4)dx$

$\int_{0}^{3}-x^{2}+x-6dx$

$[-\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+6x]_{0}^{3}$

= $\frac{27}{2}ft^{2}$

So $V = 2ft(\frac{27}{2}ft^{2})= 27ft^{3}$

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Example Question #102 : Area

Find the dot product of the two vectors.

$<9,6,4>\cdot<3,1,6>$

Possible Answers:

None of these

Correct answer:

Explanation:

The dot product of two vectors can be found by multiplying the first element of one by the first element in the other and adding it to the product of the second elements of each and so on. Using this method the dot product becomes

$<9,6,4>\cdot<3,1,6>=9\cdot3+6\cdot1+4\cdot6$

=27+6+24=57

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Example Question #103 : Area

Find the dot product of <2,1,2> and <5,5,5>

Possible Answers:

None of these

Correct answer:

Explanation:

$<2,1,2>\cdot<5,5,5>=2\cdot5+1\cdot5+2\cdot5$

=10+5+10=25

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Example Question #105 : How To Find Area Of A Region

The following problem asks for the development of a maximization problem.

With 64 inches of string, you are supposed to find the most effective use of your yarn by creating a four-sided object with the most area possible. This can be solved using basic calculus, which of the following shows the correct two equations necessary to solve this equation?

( and represent different sides of the object while represents the surface area inside the object)

Possible Answers:

2a+2b=A

ab=64

a+b=A

ab=64

4ab=A

2a+2b=64

ab=A

a+b=64

ab=A

Correct answer:

2a+2b=64

ab=A

Explanation:

If we look at the object, we will notice that we have four sides that are added up to give us our total length of string which is . This can be expressed as 2a+2b=64 . We also know that if we multiply the sides and , we will find the area, which we are trying to maximize. This leads us with our answer, and ab=A . Can you find the length of each side?

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Example Question #106 : How To Find Area Of A Region

Find the area bounded by $y = \sqrt{x}$ and y = x^3 on the interval [0,1] .

Possible Answers:

$\frac{7}{12}$

$\frac{2}{3}$

$\frac{5}{12}$

$\frac{1}{4}$

Correct answer:

$\frac{5}{12}$

Explanation:

To find the area between these two curves, we must start by determining if one function is greater than the other function over the entire interval. Because $\sqrt{x} > x^3$ over the interval [0,1] , we can simply subtract the area under y = x^3 from the area under $y = \sqrt{x}$ . Set up the integration as follows:

$\int_{0}^{1}\sqrt{x} - x^3$

To evaluate this integral, follow the power rule for integrals:

$\int x^n = \frac{x^{n+1}}{n+1}$

Realizing that $\sqrt x = x^{\frac{1}{2}}$ , the integral evaluates to:

$(\frac{2\cdot 1^{3/2}}{3} - \frac{1^4}{4}) - (\frac{2\cdot0^{3/2}}{3} - \frac{0^4}{4})$

$\frac{2}{3} - \frac{1}{4}$

$\frac{8}{12} - \frac{3}{12}$

$\frac{5}{12}$

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Example Question #107 : How To Find Area Of A Region

Find the area underneath the curve to the x-axis of the function on the interval [0,15]

f(x)=50 .

Possible Answers:

750 square units

600 square units

1000 square units

500 square units

Correct answer:

750 square units

Explanation:

In order to find the area underneath the curve to the x-axis on the interval [a,b] we must solve the integral

$\int_{a}^{b} \left | f(x)\right |\, dx$ .

Because the function is always positive on the interval [0,15] we solve the integral

$\int_{0}^{15} 50 \, dx$ .

When taking the integral we apply the inverse power rule which states

$\int x^n = \frac{x^{n+1}}{n+1}$ .

As such

$\int_{0}^{15} 50 \, dx = 50x|_0^{15}$ .

And by the corollary of the first Fundamental Theorem of Calculus

$50x|_0^{15}=50*15-50*0=750$ .

As such the area is

750 square units.

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Example Question #4091 : Calculus

Find the area underneath the curve to the x-axis of the function on the interval [0,2]

f(x)=x^4 .

Possible Answers:

$\frac{32}{5}$ square units

square units

$\frac{15}{2}$ square units

Correct answer:

$\frac{32}{5}$ square units

Explanation:

In order to find the area underneath the curve to the x-axis on the interval [a,b] we must solve the integral

$\int_{a}^{b} \left | f(x)\right |\, dx$ .

Because the function is always positive on the interval [0,2] we solve the integral

$\int_{0}^{15} x^4 \, dx$ .

When taking the integral we apply the inverse power rule which states

$\int x^n = \frac{x^{n+1}}{n+1}$ .

As such

$\int_{0}^{2} x^4 \, dx = \frac{x^5}{5}|_0^{2}$ .

And by the corollary of the first Fundamental Theorem of Calculus

$\frac{x^5}{5}|_0^{2}=\frac{2^5}{5}-0=\frac{32}{5}$ .

As such the area is

$\frac{32}{5}$ square units.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #176 : Regions

Example Question #177 : Regions

Example Question #101 : Area

Example Question #102 : How To Find Area Of A Region

Example Question #102 : Area

Example Question #103 : Area

Example Question #105 : How To Find Area Of A Region

Example Question #106 : How To Find Area Of A Region

Example Question #107 : How To Find Area Of A Region

Example Question #4091 : Calculus

All Calculus 1 Resources

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