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Calculus 1 : Area

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Example Question #1 : How To Find Area Of A Region

What is the average value of the function f(x) = 12x³ + 15x + 5 on the interval [3, 6]?

Possible Answers:

1350.2

1542

771

895.67

1302.5

Correct answer:

1302.5

Explanation:

To find the average value, we must take the integral of f(x) between 3 and 6 and then multiply it by 1/(6 – 3) = 1/3.

The indefinite form of the integral is: 3x⁴ + 7.5x² + 5x

The integral from 3 to 6 is therefore: (3(6)⁴ + 7.5(6)² + 5(6)) - (3(3)⁴ + 7.5(3)² + 5(3)) = (3888 + 270 + 30) – (243 + 22.5 + 15) = 3907.5

The average value is 3907.5/3 = 1302.5

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Example Question #2 : Area

Find the dot product of a = <2,2,-1> and b = <5,-3,2>.

Possible Answers:

-3

-2

Correct answer: 2

Explanation:

To find the dot product, we multiply the individual corresponding components and add.

Here, the dot product is found by:

2 * 5 + 2 * (-3) + (-1) * 2 = 2.

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Example Question #3 : Area

Find the area of the region enclosed by the parabola y = 2 - x^2 and the line y=-x .

Possible Answers:

$-\frac{7}{6}$

$\frac{11}{2}$

$\frac{9}{2}$

$\frac{5}{2}$

Correct answer:

$\frac{9}{2}$

Explanation:

The limits of the integration are found by solving y = 2 - x^2 and y=-x for :
2-x^2=-x
x^2-x-2=0
(x+1)(x-2)=0
x=-1, x=2

The region runs from x=-1 to x=2 . The limits of the integration are a=-1 , b=2 .
The area between the curves is:

$A = \int_{a}^{b} [f(x)-g(x)]dx = \int_{-1}^{2}[(2-x^2)-(-x))]dx$
$= \int_{1}^{2}(2+x-x^2)dx = \left [ 2x+x^2/2-x^3/3\right ]_{-1}^{2}$
$= (4+\frac{4}{2}-\frac{8}{3}) - (-2 +\frac{1}{2}+\frac{1}{3}) = \frac{9}{2}$

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Example Question #1 : Area

Find $\int \frac{3x^2-2x+1}{\sqrt{x}}dx$

Possible Answers:

$\frac{2}{3}\sqrt{x}(9x^2-10x+15)+C$

$\frac{2}{15}\sqrt{x}(9x^2-10x+15)+C$

$\frac{2}{15}x(9x^2-10x+15)+C$

$\frac{1}{15}\sqrt{x}(9x^2-10x+15)+C$

Correct answer:

$\frac{2}{15}\sqrt{x}(9x^2-10x+15)+C$

Explanation:

$\int \frac{3x^2-2x+1}{\sqrt{x}}dx=\int (3x^{3/2}-2x^{1/2}+x^{-1/2})dx \Rightarrow$

$3(\frac{2}{5})x^{5/2}-2(\frac{2}{3})x^{3/2}+2x^{1/2}+C = \frac{6}{5}x^{5/2}-\frac{4}{3}x^{3/2}+2x^{1/2}+C$

$=\frac{2}{5}\sqrt{x}(9x^2-10x+15)+C$

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Example Question #5 : Area

Find $\int \frac{1}{sec(x)}dx.$

Possible Answers:

cos(x)+C

sin(x)+C

-sin(x)+C

-cos(x)+C

Correct answer:

sin(x)+C

Explanation:

$\int \frac{1}{sec(c)}dx=\int cos(x)dx=sin(x)+C$

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Example Question #6 : Area

What is the area of the space below y = x and above y = x^2

Possible Answers:

$\frac{1}{2}$

$\frac{1}{5}$

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{6}$

Explanation:

y = x is only above y = x^2 over the interval (0,1) . Areas are given by the definite integral of each function $a_1 = \int_{0}^{1}xdx$ and $a_2 = \int_{0}^{1}x^2dx$

The area between the curves is found by subtracting the area between each curve and the -axis from each other. For y = x this area is $\frac{1}{2}$ and for y = x^2 the area is $\frac{1}{3}$ giving an area between curves of $\frac{1}{6}$

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Example Question #7 : Area

What is the area below y = -x^2+4 and above the -axis?

Possible Answers:

$\frac{32}{3}$

$\frac{16}{3}$

$\frac{8}{3}$

Correct answer:

$\frac{32}{3}$

Explanation:

To find the area below a curve, you must find the definite integral of the function. In this case the limits of integration are where the original function intercepts the -axis at and . So you must find $\int_{-2}^{2}(-x^2+4)dx$ which is $-\frac{1}{3}x^3+4 x$ evaluated from to . This gives an answer of $\frac{32}{3}$

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Example Question #8 : Area

Find the area between the curves y=2x and y=x^2 .

Possible Answers:

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{5}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

To solve this problem, we first need to find the point where the two equations are equal. Doing this we find that

$2x=x^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0$ .

From this, we see that the two graphs are equal at x=0 and x=2 . We also know that for 0<x<2 , y=2x is greater than y=x^2 .

So to find the area between these curves we need to evaluate the integral $\int_{0}^{2} (2x-x^2)dx$ .

The solution to the integral is

$\int(2x-x^2)dx=x^2-\frac{1}{3}x^3$ .

Evaluating this at x=2 and x=0 we get

$(2*2-\frac{1}{3}2^3)-(0-0)=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}.$

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Example Question #9 : Area

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

Possible Answers:

$\frac{1}{2}$

$\frac{3}{2}$

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 #10 : Area

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

Possible Answers:

1/12

1/6

-1/3

1/3

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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Calculus 1 : Area

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : How To Find Area Of A Region

Example Question #2 : Area

Example Question #3 : Area

Example Question #1 : Area

Example Question #5 : Area

Example Question #6 : Area

Example Question #7 : Area

Example Question #8 : Area

Example Question #9 : Area

Example Question #10 : Area

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