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Calculus 1 : Regions

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Example Questions

Example Question #101 : Area

Find the area of the region bounded by the function $\small f(x)=4-4e^{-2x}$ and the -axis over the interval $\small x=[0,2]$ .

Possible Answers:

$\small 8+2e^{-4}$

$\small 10-2e^{-4}$

$\small 10+2e^{-4}$

$\small 8-2e^{-4}$

$\small 6+2e^{-4}$

Correct answer:

$\small 6+2e^{-4}$

Explanation:

To find the area of the region between $\small f(x)=4-4e^{-2x}$ and the x-axis, integrate the function over the specified interval:

For this particular function use the following rules to integrate.

$\int x^n=\frac{x^{n+1}}{n+1}$ and $\int e^u=\frac{1}{\frac{du}{dx}}e^u$

Therefore we get,

$\small A=\int_0^2(4-4e^{-2x})dx$

$\small A=4x+2e^{-2x}|_0^2$

$\small A=(8+2e^{-4})-(4(0)+2e^{-2(0)})$

$\small A=8+2e^{-4}-2$

$\small \small A=6+2e^{-4}$ .

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Example Question #102 : How To Find Area Of A Region

A construction company is pouring concrete pads for an abstract art-themed promenade on a college campus. One slab's overhead view is given by region contained between the functions f(x)=x+2 and $g(x)=x^{2}-4$ and to the right of the -axis, where gives the size in feet. If the slab is two feet deep, what is the volume occupied by the concrete slab?

Possible Answers:

$\frac{68}{3}ft^{3}$

$\frac{27}{2}ft^{3}$

$54 ft^{3}$

$27 ft^{3}$

Correct answer:

$27 ft^{3}$

Explanation:

Since the depth of the slab is constant, we know it's volume is given by:

V= Ah =2A

The area is, of course given by an integral. To calculate it, we must know where the two functions intersect:

$x+2 = (x^{2}-4)$

$x^{2}-x-6 = 0$

Which has roots x=-2 and x=3 . Since we are concerned with area to the right of the y-axis, we discard the negative root. So the area is given by:

$A=\int_{0}^{3}x+2-(x^{2}-4)dx$

$\int_{0}^{3}-x^{2}+x-6dx$

$[-\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+6x]_{0}^{3}$

= $\frac{27}{2}ft^{2}$

So $V = 2ft(\frac{27}{2}ft^{2})= 27ft^{3}$

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Example Question #101 : How To Find Area Of A Region

Find the dot product of the two vectors.

$<9,6,4>\cdot<3,1,6>$

Possible Answers:

None of these

Correct answer:

Explanation:

The dot product of two vectors can be found by multiplying the first element of one by the first element in the other and adding it to the product of the second elements of each and so on. Using this method the dot product becomes

$<9,6,4>\cdot<3,1,6>=9\cdot3+6\cdot1+4\cdot6$

=27+6+24=57

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Example Question #101 : How To Find Area Of A Region

Find the dot product of <2,1,2> and <5,5,5>

Possible Answers:

None of these

Correct answer:

Explanation:

$<2,1,2>\cdot<5,5,5>=2\cdot5+1\cdot5+2\cdot5$

=10+5+10=25

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Example Question #105 : How To Find Area Of A Region

The following problem asks for the development of a maximization problem.

With 64 inches of string, you are supposed to find the most effective use of your yarn by creating a four-sided object with the most area possible. This can be solved using basic calculus, which of the following shows the correct two equations necessary to solve this equation?

( and represent different sides of the object while represents the surface area inside the object)

Possible Answers:

2a+2b=A

ab=64

a+b=A

ab=64

4ab=A

2a+2b=64

ab=A

a+b=64

ab=A

Correct answer:

2a+2b=64

ab=A

Explanation:

If we look at the object, we will notice that we have four sides that are added up to give us our total length of string which is . This can be expressed as 2a+2b=64 . We also know that if we multiply the sides and , we will find the area, which we are trying to maximize. This leads us with our answer, and ab=A . Can you find the length of each side?

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Example Question #106 : How To Find Area Of A Region

Find the area bounded by $y = \sqrt{x}$ and y = x^3 on the interval [0,1] .

Possible Answers:

$\frac{7}{12}$

$\frac{2}{3}$

$\frac{5}{12}$

$\frac{1}{4}$

Correct answer:

$\frac{5}{12}$

Explanation:

To find the area between these two curves, we must start by determining if one function is greater than the other function over the entire interval. Because $\sqrt{x} > x^3$ over the interval [0,1] , we can simply subtract the area under y = x^3 from the area under $y = \sqrt{x}$ . Set up the integration as follows:

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

To evaluate this integral, follow the power rule for integrals:

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

Realizing that $\sqrt x = x^{\frac{1}{2}}$ , the integral evaluates to:

$(\frac{2\cdot 1^{3/2}}{3} - \frac{1^4}{4}) - (\frac{2\cdot0^{3/2}}{3} - \frac{0^4}{4})$

$\frac{2}{3} - \frac{1}{4}$

$\frac{8}{12} - \frac{3}{12}$

$\frac{5}{12}$

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Example Question #4091 : Calculus

Find the area underneath the curve to the x-axis of the function on the interval [0,15]

f(x)=50 .

Possible Answers:

750 square units

600 square units

1000 square units

500 square units

Correct answer:

750 square units

Explanation:

In order to find the area underneath the curve to the x-axis on the interval [a,b] we must solve the integral

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

Because the function is always positive on the interval [0,15] we solve the integral

$\int_{0}^{15} 50 \, dx$ .

When taking the integral we apply the inverse power rule which states

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

As such

$\int_{0}^{15} 50 \, dx = 50x|_0^{15}$ .

And by the corollary of the first Fundamental Theorem of Calculus

$50x|_0^{15}=50*15-50*0=750$ .

As such the area is

750 square units.

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Example Question #102 : How To Find Area Of A Region

Find the area underneath the curve to the x-axis of the function on the interval [0,2]

f(x)=x^4 .

Possible Answers:

square units

$\frac{32}{5}$ square units

$\frac{15}{2}$ square units

square units

Correct answer:

$\frac{32}{5}$ square units

Explanation:

In order to find the area underneath the curve to the x-axis on the interval [a,b] we must solve the integral

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

Because the function is always positive on the interval [0,2] we solve the integral

$\int_{0}^{15} x^4 \, dx$ .

When taking the integral we apply the inverse power rule which states

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

As such

$\int_{0}^{2} x^4 \, dx = \frac{x^5}{5}|_0^{2}$ .

And by the corollary of the first Fundamental Theorem of Calculus

$\frac{x^5}{5}|_0^{2}=\frac{2^5}{5}-0=\frac{32}{5}$ .

As such the area is

$\frac{32}{5}$ square units.

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Example Question #101 : How To Find Area Of A Region

Find the area underneath the curve to the x-axis of the function on the interval [0,1]

f(x)=e^x .

Possible Answers:

square units

square unit

e-1 square units

Correct answer:

e-1 square units

Explanation:

In order to find the area underneath the curve to the x-axis on the interval [a,b] we must solve the integral

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

Because the function is always positive on the interval [0,1] we solve the integral

$\int_{0}^{1} e^x \, dx$ .

Because the antiderivative of the exponential function is the exponential function itself, we obtain

$\int_{0}^{1} e^x \, dx = e^x|_0^1$ .

And by the corollary of the first Fundamental Theorem of Calculus

e^x|_0^1 = e^1-e^0=e-1 .

As such the area is

e-1 square units.

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Example Question #101 : How To Find Area Of A Region

Evaluate the following integral to find the area of the region bound by the function and the given limits.

$\int_{0}^{5}x^4-6x^2+8x+6dx$

Possible Answers:

505 units^2

5 units^2

275 units^2

475 units^2

Correct answer:

505 units^2

Explanation:

Evaluate the following integral to find the area of the region bound by the function and the given limits.

$\int_{0}^{5}x^4-6x^2+8x+6dx$

Begin by recalling the integration rule for polynomials. All we need to do to integrate a term of a polynomial is raise the exponent by 1 and divide by the new number.

So, we go from this

$\int_{0}^{5}x^4-6x^2+8x+6dx$

To this

$\frac{x^5}{5}-\frac{6x^3}{3}+\frac{8x^2}{2}+6x+c|_0^5$

We now have successfully integrate our function, but we still must evaluate it with the given limits. To do so, simply plug in our lower and upper limits and find the difference between them. Note, this is step is simplified by the fact that our lower limit is 0, because plugging zero into our function will just yield "c"

So...

$\frac{5^5}{5}-\frac{6(5)^3}{3}+\frac{8(5)^2}{2}+6(5)+c-c$

5^4-2*5^3+4*5^2+30=505

Making our answer

505 units^2

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Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #101 : Area

Example Question #102 : How To Find Area Of A Region

Example Question #101 : How To Find Area Of A Region

Example Question #101 : How To Find Area Of A Region

Example Question #105 : How To Find Area Of A Region

Example Question #106 : How To Find Area Of A Region

Example Question #4091 : Calculus

Example Question #102 : How To Find Area Of A Region

Example Question #101 : How To Find Area Of A Region

Example Question #101 : How To Find Area Of A Region

All Calculus 1 Resources

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