Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #2991 : Functions

Find the area beneath the curve defined by the following function

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

between \displaystyle x = 0 to \displaystyle \pi.

Possible Answers:

\displaystyle \pi ^3 - 2

\displaystyle 6\pi ^3 + 2

\displaystyle 6\pi ^3 - 2

\displaystyle \pi ^3 + 2

Correct answer:

\displaystyle \pi ^3 - 2

Explanation:

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

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

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

\displaystyle x^3 + cos(x) from \displaystyle 0 to \displaystyle \pi.

or

\displaystyle (\pi ^3 + cos(\pi )) - (0^3 + cos(0)) which simplifies to  \displaystyle \pi3 \displaystyle - 2

 

Example Question #2992 : Functions

A curve is defined by the function:

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

Find the area between the curve and the \displaystyle x-axis on the interval of \displaystyle x = 0 to \displaystyle 1.

Possible Answers:

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

\displaystyle 2e^2-4

\displaystyle -4

\displaystyle \frac{35}{12}

Correct answer:

\displaystyle \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:

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

\displaystyle 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:

\displaystyle 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}

 

 

 

Example Question #2993 : Functions

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

Possible Answers:

\displaystyle 106

\displaystyle 160

\displaystyle 356

\displaystyle 64

\displaystyle 157

Correct answer:

\displaystyle 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: 

\displaystyle \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,

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

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

2. \displaystyle \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 \displaystyle 160 after rounding.

Example Question #111 : Regions

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

Possible Answers:

\displaystyle 164

\displaystyle 157

\displaystyle 143

\displaystyle 176

\displaystyle 152

Correct answer:

\displaystyle 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:

 \displaystyle \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,

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

\displaystyle \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.

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

Example Question #31 : Area

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

Possible Answers:

\displaystyle 9

\displaystyle 8

\displaystyle 11

\displaystyle 12

\displaystyle 10

Correct answer:

\displaystyle 11

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:

\displaystyle \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,

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

\displaystyle \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.

\displaystyle \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.

 

 

 

Example Question #113 : Regions

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

Possible Answers:

\displaystyle 2 + \pi ^{2}

\displaystyle 1 + \pi^{2}

\displaystyle 2 + \pi

\displaystyle \pi

\displaystyle 0

Correct answer:

\displaystyle 2 + \pi ^{2}

Explanation:

Written in words, solve:

\displaystyle \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,

\displaystyle \int x^n=\frac{x^{n+1}}{n+1} to the term \displaystyle 2x and recalling the integral of \displaystyle sin(x) is \displaystyle -cos(x) we find,

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

2. Plug in \displaystyle \pi and \displaystyle 0 for \displaystyle x and then take the difference.

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

note:

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

Example Question #2991 : Functions

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

Possible Answers:

\displaystyle e^{\pi-1}

\displaystyle e^{\pi} + 1

\displaystyle e^{\pi}-1

\displaystyle e^{\pi}

\displaystyle 1

Correct answer:

\displaystyle 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:

\displaystyle \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 \displaystyle cos(x) is \displaystyle sin(x) and that the integral of \displaystyle e^x is itself, we find that,

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

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

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

 

Example Question #115 : Regions

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

Possible Answers:

\displaystyle 6\pi^{2}

\displaystyle 0

\displaystyle 6\pi

\displaystyle \pi-6

\displaystyle \pi + 6

Correct answer:

\displaystyle 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:

\displaystyle \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,

\displaystyle \int x^n=\frac{x^{n+1}}{n+1} to the term \displaystyle 6 and recalling the integral of \displaystyle cos(x) is \displaystyle sin(x) we find,

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

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

Example Question #116 : Regions

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

Possible Answers:

\displaystyle 0

\displaystyle 4\pi^{2}

\displaystyle 2\pi^{2}

\displaystyle 2\pi

\displaystyle 4\displaystyle \pi

Correct answer:

\displaystyle 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:

\displaystyle \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,

\displaystyle \int x^n=\frac{x^{n+1}}{n+1} to the term \displaystyle 2x and recalling the integral of \displaystyle cos(x) is \displaystyle sin(x) we find,

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

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

Example Question #117 : Regions

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

Possible Answers:

\displaystyle e^2 + 1

\displaystyle 2e^2 - 4

\displaystyle e^2 - 4

\displaystyle e^2

\displaystyle e^2 -1

Correct answer:

\displaystyle 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:

\displaystyle \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,

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

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

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

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