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Calculus 1 : Functions

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Example Questions

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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Example Question #41 : Area

Find the area under the curve |4-x^2| from x=-2 to x=3 .

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 as the following:

$\int_{-2}^{3} |4-x^2|dx$

There are 2 things to note.

1. This function is positive only from x=-2 to x=2 . From x=2 to , this function is negative.

2. This function is not differentiable at x=2 because of the absolute value.

Therefore, this function needs to be broken up into 2 pieces.

$\int_{-2}^{2} |4-x^2|dx + \int_{2}^{3} x^2-4dx$

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 two values.

1. Using the power rule which states,

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

$\int 4 - x^2dx = 4x - \frac{x^3}{3} +constant$

$\int x^2-4dx =\frac{x^3}{3} -4x +constant$ .

2. Plug in the respective upper and lower limits for x and then take the difference.

$4x - \frac{x^3}{3}|_{-2}^{2} = 8-\frac{8}{3}-\left(-8 + \frac{8}{3}\right) = 16 - \frac{16}{3} = \frac{32}{3}$

$\frac{x^3}{3} - 4x|_{2}^3 = \frac{27}{3} - 12 -\left(\frac{8}{3}-8\right) = 9 - 12 + 8 - \frac{8}{3} = 5 - \frac{8}{3}$

Now take the sum of the integrals of the two pieces.

$\frac{32}{3}+ 5 - \frac{8}{3} = 5 + \frac{24}{3} = 5 + 8 = 13$

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Example Question #41 : Area

Find the area under $y=\sin{x}$ where $0<x<\pi/2$ .

Possible Answers:

$\frac{\pi}{2}$

$\pi$

Correct answer:

Explanation:

Find the area under $y=\sin{x}$ where $0<x<\pi/2$ .

Take the integral of $f(x)=\sin{x}$ :

$F(x)=-\cos{x}$

Next, evaluate it on the interval:

$F(\pi/2)-F(0)$

$(-\cos(\pi/2))-(-\cos(0))$

(-0)-(-1)

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Example Question #4031 : Calculus

Find the the area of the region defined by $f(x)=\ln(x)$ , the -axis, and the line x=e .

Possible Answers:

e-1

e+1

-2e+1

Correct answer:

Explanation:

To find this area, you can either take the integral of ln(x) from 1 to , or you can redefine the integral in terms of and solve that way, as the integral of the natural log is more difficult to find.

To do this, rewrite y=ln(x) as e^y=x and then redefine the bounds of the integral, so instead of integrating on (1,e) you can integrate on (0,1) .

The final integral is

$\int_0^1 e^y dy= e-1$ .

This however, is the area between the curve and the y-axis. To find the remaining area, subtract this from the area from to the -axis on (0,1) , which has an area of .

Thus, the final expression resulting in the answer is,

e-(e-1)=1 .

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Example Question #121 : Regions

What is the area of the region bounded by $\small \small y=x^2, x=5$ and $\small y=0$ ?

Possible Answers:

$\small 125/2$

$\small 125$

$\small 125/3$

$\small 0$

Correct answer:

$\small 125/3$

Explanation:

To find the area under the curve, we need to perform a definite integral. Essentially, this integral will be summing up all the infinitesimally small rectangles that make up the region. The entire region is in the first quadrant, so we don't have to worry about splitting our region up.

When we take the integral we will need to use,

$\int x^n=\frac{x^{n+1}}{n+1}$ then plug in the upper and lower bounds into the function and take the difference.

Therefore,

$\small \int_{0}^{5}(x^2)dx=(5)^3/3-0^3/3=125/3$

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Example Question #42 : Area

Let $f(x)=2cos(\frac{\pi x}{4})$

Find the area enclosed under f(x) on the interval from x=0 to x=2 .

Possible Answers:

$\frac {8}{\pi}$

$\sqrt{2}$

$2\pi$

Correct answer:

$\frac {8}{\pi}$

Explanation:

To find the area under the curve, evaluate

$\int_{0}^{2}f(x)dx$ .

$\int_{0}^{2} 2cos\left(\frac{\pi x}{4}\right)dx$

Remember the integral of cos(x) is $sin(u)\cdot \frac{1}{\frac{du}{dx}}$ .

Integrating yields:

$= \frac{8}{\pi}sin\left(\frac{\pi x}{4}\right)|_{0}^{2}$

Now, evaluate:

$= \frac{8}{\pi}sin\left(\frac{\pi\cdot 2 }{4}\right) - \frac{8}{\pi}sin(0) = \frac{8}{\pi}sin\left(\frac{\pi }{2}\right)-0$

$=\frac{8}{\pi}$

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Example Question #46 : How To Find Area Of A Region

Find the area of the region bounded by the functions f(x)=x+1 and g(x)=x^2+1 .

Possible Answers:

$\frac{3}{2}$

$\frac{7}{6}$

$\frac{1}{6}$

$\frac{9}{2}$

$\frac{5}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

If $f(x)\geq g(x)$ on the interval $a\leq x \leq b$ , then the area bounded by the two functions is given by

$\int_a^b (f(x)-g(x))dx$ .

To find the area bounded by x+1 and x^2+1 , we must first find when they intersect.

Setting the functions equal x+1=x^2+1 , we get 0=x^2-x=x(x-1) .

Therefore the two functions intersect at x=0 and x=1 and we wish to find the area of the region bounded by f(x) and g(x) between these two points. By plotting the two functions, we see that on this interval, $x+1\geq x^2+1$ .

Therefore, the area of the region bounded by these functions is given by

$\int_0^1 \left((x+1)-(x^2+1) \right )dx =\int_0^1 \left(-x^2+x \right )dx$

$=\left[-\frac{x^3}{3}+\frac{x^2}{2} \right ]_0^1=\left(-\frac{1^3}{3}+\frac{1^2}{2} \right )-\left( -\frac{0^3}{3}+\frac{0^2}{2}\right )$

$=-\frac{1}{3}+\frac{1}{2}=\frac{1}{6}$ .

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Example Question #43 : Area

Find the area of the region bounded by the the graph and the -axis, given

$f(x)=\sqrt{x}$ on the interval [0,16] .

Possible Answers:

128 square units

$\frac{128}{3}$ square units

$\frac{64}{3}$ square units

square units

Correct answer:

$\frac{128}{3}$ square units

Explanation:

The area for the region bounded by the graph of f(x) and the x-axis on the interval [a,b] is given as

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

Because f(x)>0 on the interval [0,16] the area is

$\int_{0}^{16}f(x)\,dx=\int_{0}^{16}x^\frac{1}{2}\,dx$ .

When taking the integral, we will use the inverse power rule which states,

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

As such,

$\int_{0}^{16}x^\frac{1}{2}\,dx=\frac{x^{3/2}}{3/2}\,\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}=\frac{2}{3}x^\frac{3}{2}\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}$

And by the corollary of the First Fundamental Theorem of Calculus,

$\frac{2}{3}x^\frac{3}{2}\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}=\left[\frac{2}{3}(16)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(0)^{\frac{3}{2}}\right]$ .

Hence the area is

$\frac{128}{3}$ square units.

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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

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

Example Question #4031 : Calculus

Example Question #41 : Area

Example Question #41 : Area

Example Question #4031 : Calculus

Example Question #121 : Regions

Example Question #42 : Area

Example Question #46 : How To Find Area Of A Region

Example Question #43 : Area

All Calculus 1 Resources

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