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Calculus 1 : Functions

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Example Questions

Example Question #2991 : Functions

Find the area beneath the curve defined by the following function

f(x) = 3x^2 - sin (x)

between x = 0 to $\pi$ .

Possible Answers:

$\pi ^3 + 2$

$6\pi ^3 - 2$

$6\pi ^3 + 2$

$\pi ^3 - 2$

Correct answer:

$\pi ^3 - 2$

Explanation:

To find the area underneath a curve within a given integral, integrate the curve between the upper and lower limits:

$\int_{0}^{\pi } (3x^{2} -sin(x)) dx$

The integral of 3x² is x³ and the integral of - sin(x) is cos(x), giving:

x^3 + cos(x) from to $\pi$ .

$(\pi ^3 + cos(\pi )) - (0^3 + cos(0))$ which simplifies to $\pi$ ³

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Example Question #29 : How To Find Area Of A Region

A curve is defined by the function:

$y=3x^3 - 4x^2 +8x +e^{2x}$

Find the area between the curve and the -axis on the interval of x = 0 to .

Possible Answers:

2e^2-4

$\frac{35}{12}$

$\frac{35+6e^2}{12}$

Correct answer:

$\frac{35+6e^2}{12}$

Explanation:

To find the area between the curve and the x-axis, we will need to integrate the curve with respect to x:

$y=3x^3 - 4x^2 +8x +e^{2x}$

$Y = \frac{3x^4}{4}-\frac{4x^3}{3}+4x^2+\frac{e^{2x}}{2}$ from x = 0 to 1

Introducing are upper and lower bounds, we can find the area to be:

$Area =\left(\frac{3}{4}-\frac{4}{3}+4+\frac{e^2}{2}\right) - \left(0-0+0+\frac{1}{2}\right)=\frac{35+6e^2}{12}$

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Example Question #2991 : Functions

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

Possible Answers:

157

356

106

160

Correct answer:

160

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:

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

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.

1. Using the power rule which states,

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

$\int 5x^{2} + e^{x} dx = \frac{5}{3}x^{3} + e^{x} + C$ .

2. $\left( \frac{5}{3}4^{3} + e^{4} \right) - \left( \frac{5}{3}0^{3} + e^{0} \right) = \frac{320}{3} + e^4 - 1 = 160.26$ , which is 160 after rounding.

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Example Question #2992 : Functions

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

Possible Answers:

152

176

164

143

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 #2993 : Functions

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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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2991 : Functions

Example Question #29 : How To Find Area Of A Region

Example Question #2991 : Functions

Example Question #2992 : Functions

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

Example Question #31 : Area

Example Question #2993 : Functions

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

Example Question #111 : Regions

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

All Calculus 1 Resources

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