Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #101 : Area

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the area of the region between \displaystyle \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.

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

Therefore we get,

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

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

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

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

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

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  \displaystyle f(x)=x+2 and  \displaystyle g(x)=x^{2}-4 and to the right of the \displaystyle y-axis, where \displaystyle x gives the size in feet. If the slab is two feet deep, what is the volume occupied by the concrete slab?

Possible Answers:

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

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

\displaystyle 54 ft^{3}

\displaystyle 27 ft^{3}

Correct answer:

\displaystyle 27 ft^{3}

Explanation:

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

\displaystyle V= Ah =2A

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

 

\displaystyle x+2 = (x^{2}-4) 

\displaystyle x^{2}-x-6 = 0

 

Which has roots \displaystyle x=-2 and \displaystyle 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:

 

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

 

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

 

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

 

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

 

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

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

Find the dot product of the two vectors.

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

Possible Answers:

\displaystyle 50

\displaystyle 0

\displaystyle 24

\displaystyle 57

None of these

Correct answer:

\displaystyle 57

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 

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

\displaystyle =27+6+24=57

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

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

Possible Answers:

\displaystyle 0

None of these

\displaystyle 6

\displaystyle 20

\displaystyle 25

Correct answer:

\displaystyle 25

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 

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

\displaystyle =10+5+10=25

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?

 

(\displaystyle a and \displaystyle b represent different sides of the object while \displaystyle A represents the surface area inside the object)

Possible Answers:

\displaystyle 2a+2b=A

\displaystyle ab=64

\displaystyle a+b=A

\displaystyle ab=64

\displaystyle 4ab=A

\displaystyle 2a+2b=64

\displaystyle 2a+2b=64

\displaystyle ab=A

\displaystyle a+b=64

\displaystyle ab=A

Correct answer:

\displaystyle 2a+2b=64

\displaystyle 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 \displaystyle 64.  This can be expressed as \displaystyle 2a+2b=64.  We also know that if we multiply the sides \displaystyle a and \displaystyle b, we will find the area, which we are trying to maximize.  This leads us with our answer, \displaystyle 2a+2b=64 and \displaystyle ab=A. Can you find the length of each side?

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

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

Possible Answers:

\displaystyle \frac{7}{12}

\displaystyle \frac{2}{3}

\displaystyle 0

\displaystyle \frac{5}{12}

\displaystyle \frac{1}{4}

Correct answer:

\displaystyle \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 \displaystyle \sqrt{x} > x^3 over the interval \displaystyle [0,1], we can simply subtract the area under \displaystyle y = x^3 from the area under \displaystyle y = \sqrt{x}. Set up the integration as follows:

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

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

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

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

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

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

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

\displaystyle \frac{5}{12}

Example Question #4091 : Calculus

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

if

\displaystyle f(x)=50.

Possible Answers:

\displaystyle 750 square units

\displaystyle 600 square units

\displaystyle 1000 square units

\displaystyle 500 square units

Correct answer:

\displaystyle 750 square units

Explanation:

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

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

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

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

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

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

As such

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

And by the corollary of the first Fundamental Theorem of Calculus

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

As such the area is

\displaystyle 750 square units.

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 \displaystyle [0,2]

if

\displaystyle f(x)=x^4.

Possible Answers:

\displaystyle 6 square units

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

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

\displaystyle 5 square units

Correct answer:

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

Explanation:

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

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

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

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

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

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

As such

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

And by the corollary of the first Fundamental Theorem of Calculus

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

As such the area is

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

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 \displaystyle [0,1]

if

\displaystyle f(x)=e^x.

Possible Answers:

\displaystyle e square units

\displaystyle e-1 square units

\displaystyle -e square units

\displaystyle 1 square unit

Correct answer:

\displaystyle e-1 square units

Explanation:

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

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

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

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

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

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

And by the corollary of the first Fundamental Theorem of Calculus

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

As such the area is

\displaystyle e-1 square units.

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.

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

Possible Answers:

\displaystyle 505 units^2

\displaystyle 5 units^2

\displaystyle 275 units^2

\displaystyle 475 units^2

Correct answer:

\displaystyle 505 units^2

Explanation:

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

\displaystyle \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

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

To this

\displaystyle \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...

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

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

Making our answer 

\displaystyle 505 units^2

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