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Calculus 1 : Area

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Example Questions

Example Question #4021 : Calculus

Find the area under the curve $x^{5} + 2x^{2} + 4$ from x = 0 to x = 3 , rounded to the nearest integer.

Possible Answers:

143

152

164

176

157

Correct answer:

152

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_{0}^{3} x^{5} + 2x^{2} + 4dx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the 2 values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to each term we find,

$\int x^5 + 2x^2 + 4dx = \frac{x^6}{6} + \left(\frac{2}{3}\right)x^{3} + 4x + constant$

2. Plug in 3 and 0 for x and then take the difference.

$3^{6} + \left(\frac{2}{3}\right)\cdot 3^{3} + 4\cdot3 - 0$ = 151.50 , which rounds up to 152 .

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Example Question #32 : How To Find Area Of A Region

Find the area under the curve $(x - 3)^{2} + 2x$ from x = 1 to x= 3 , rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

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

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to each term we find,

$\int (x-3)^2 + 2xdx = \frac{x^{3}}{3} - 2x^{2} + 9x$ .

2. Plug in 4 and 1 for x and then take the difference.

$\left(\frac{x^3}{3} - 2\cdot3^{2} + 9\cdot3\right) - \left(\frac{1}{3} - 2\cdot1^{2} + 9\right) = 10.667$

The answer is 11 because 10.667 rounds up.

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Example Question #31 : Area

Find the area of the curve sin(x) + 2x from x = 0 to x = $\pi$

Possible Answers:

$\pi$

$1 + \pi^{2}$

$2 + \pi$

$2 + \pi ^{2}$

Correct answer:

$2 + \pi ^{2}$

Explanation:

Written in words, solve:

$\int_{0}^{\pi} sin(x) + 2x dx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to the term and recalling the integral of sin(x) is -cos(x) we find,

$\int sin(x) + 2xdx = x^{2} - cos(x)$ .

2. Plug in $\pi$ and for and then take the difference.

$\pi^{2} - cos(\pi) -(0- cos(0) )$ = $\pi^{2} + 2$

note:

$cos(0) = 1 , cos(\pi) = -1$

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Example Question #32 : Area

Find the area under the curve $cos(x) + e^{x}$ from x = 0 to x = $\pi$ .

Possible Answers:

$e^{\pi-1}$

$e^{\pi}$

$e^{\pi} + 1$

$e^{\pi}-1$

Correct answer:

$e^{\pi}-1$

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_{0}^{\pi} cos(x) + e^{x}dx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Recall that the integral of cos(x) is sin(x) and that the integral of e^x is itself, we find that,

$\int cos(x) + e^{x}dx = e^{x} + sin(x)$ .

2. $e^{\pi} + sin(\pi) - (e^{0} + sin(0))$ $= e^{\pi}-1$

note: $sin(\pi) = sin(\pi) = 0$

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Example Question #34 : How To Find Area Of A Region

Find the area under the curve cos(x) + 6 from x = 0 to x = $\pi$ .

Possible Answers:

$6\pi$

$\pi-6$

$6\pi^{2}$

$\pi + 6$

Correct answer:

$6\pi$

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_{0}^{\pi}cos(x) + 6 dx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to the term and recalling the integral of cos(x) is sin(x) we find,

$\int cos(x) + 6dx = sin(x)+6x+constant$

2. $sin(\pi)+6\pi - (sin(0) + 6\cdot0) = 6\pi$

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Example Question #111 : Regions

Find the area under the curve cos(x)+2x from x=0 to $2\pi$ .

Possible Answers:

$\pi$

$4\pi^{2}$

$2\pi$

$2\pi^{2}$

Correct answer:

$4\pi^{2}$

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_0^{2\pi}cos(x)+2xdx$

To solve:

1. Find the indefinite integral of the function

2. Plug in the upper and lower limit values and take the difference of the 2 values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to the term and recalling the integral of cos(x) is sin(x) we find,

$\int cos(x)+2xdx = x^{2}+sin(x)+constant$ .

2. $(2\pi)^{2}+sin(2\pi) - (0^2 + sin(0)) = (2\pi)^2 = 4\pi^2$

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Example Question #35 : How To Find Area Of A Region

Find the area under the curve e^2 - 2x from x=0 to x=2 .

Possible Answers:

e^2 + 1

2e^2 - 4

e^2

e^2 - 4

e^2 -1

Correct answer:

2e^2 - 4

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten into the following:

$\int_{0}^{2} e^2 - 2xdx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Using the power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ to each term we find,

$\int e^2 - 2xdx = xe^2 - x^2 + constant$ .

2. (2)e^2 - 2^2 - (0e^2 - 0^2) = 2e^2 - 4

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Example Question #36 : How To Find Area Of A Region

Find the area of the region bounded by y = x^3 and y =x on the interval 0 < x < 1 .

Possible Answers:

0.25

1.25

0.75

0.5

Correct answer:

0.25

Explanation:

Finding the area of a region can be understood as taking the integral of the difference of the two equations and it can be rewritten into the following:

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

To determine which equation gets subtracted from, try drawing the graph or testing which function is larger at a point in the interval. The function that is larger will be on top, which in this case is y=x . For example, at point x =0.2 , y=x is 0.2 while y=x^3 is 0.008 .

$\int_{0}^{1} x - x^3dx = \frac{x^2}{2} - \frac{x^4}{4} |_0^{1} = \left(\frac{1}{2} - \frac{1}{4}\right) - (0-0) = \frac{1}{4} = 0.25$

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Example Question #34 : How To Find Area Of A Region

Find the area bounded by the region y= 1-x and e^x from x = 0 to x=1 .

Possible Answers:

$e - \frac{3}{2}$

$e - \frac{1}{2}$

$e + \frac{1}{2}$

e - 1

Correct answer:

$e - \frac{3}{2}$

Explanation:

Finding the area of a region can be understood as taking the integral of the difference of the two equations and can be rewritten into the following:

$\int_{0}^{1} e^x - (1-x)dx$

To determine which equation gets subtracted from, try drawing the graph or testing which function is larger at a point in the interval. The function that is larger will be on top, which in this case is y=e^x . For example, at point x = .5 , e^x is 1.65 while y=1-x is .

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. $\int e^x - (1-x)dx = e^x + \frac{x^2}{2} - x + constant$

2. $e^1 + \frac{1}{2} - 1 - (e^0) = e + \frac{1}{2} - 1 -1 = e - \frac{3}{2}$

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Example Question #4031 : Calculus

Find the area under the curve |x^3-8| from x= -2 to x=2 .

Possible Answers:

Correct answer:

Explanation:

Finding the area under the curve can be understood as taking the integral of the equation and can be rewritten as the following:

$\int_{-2}^{2} |x^3-8|dx$

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the 2 values.

Note: Without the absolute value, this function is always negative up to x=2.

However, since we have the absolute value, the whole function is made positive. Taking away the absolute values sign is the same as multiplying the function by -1 because the whole function is negative.

1. $\int |x^3-8|dx = \int 8-x^3dx = 8x-\frac{x^4}{4} + constant$

2. Plug in 2 and -2 for x and then take the difference.

$8\cdot2 - \frac{2^4}{4} - \left(-2\cdot8 - \frac{(-2)^4}{4}\right) = 16 -4+16 + 4 = 32$

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Calculus 1 : Area

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #4021 : Calculus

Example Question #32 : How To Find Area Of A Region

Example Question #31 : Area

Example Question #32 : Area

Example Question #34 : How To Find Area Of A Region

Example Question #111 : Regions

Example Question #35 : How To Find Area Of A Region

Example Question #36 : How To Find Area Of A Region

Example Question #34 : How To Find Area Of A Region

Example Question #4031 : Calculus

All Calculus 1 Resources

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