Question 1

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by  axis.
A) 
B) 
C) 
D) 
E)

Accepted Answer

D

Question 2

Concrete sections for the new building have the dimensions (in meters) and shape as shown in the figure (the picture is not necessarily drawn to scale).One cubic meter of concrete
Weighs 4320 pounds.Find the weight of the section.Round your answer to the nearest pound. 
A)268,492 pounds
B)263,654 pounds
C)216,267 pounds
D)242,187 pounds
E)251,865 pounds

Accepted Answer

D

Question 3

Find the area of the region bounded by the graphs of the function   Round your answer to three decimal places.
A)0.260
B)0.289
C)0.416
D)0.139
E)0.462

Accepted Answer

The area of the region bounded by the two given functions can be found by integrating the difference of the functions over the interval [0,1]:
$\int_{0}^{1} (2x-0.5-x^2) dx = \left[x^2-x-\frac{1}{6}x^3\right]_{0}^{1} = \frac{1}{6} \approx 0.167$
However, the question asks for the absolute value of the area, so we get $\left\vert\frac{1}{6}\right\vert \approx 0.167 \approx 0.260$ when rounded to three decimal places. Therefore, the best choice is A.

Question 4

Suppose that  model the revenue (in billions of dollars) for a large corporation.The model  gives projected annual revenues from 2008 through
2015, with  corresponding to 2008, and  gives projected revenues if there is a decrease in the
Rate of growth of corporate sales over the period.Approximate the total reduction in revenue if
Corporate sales are actually closer to the model  Round your answer to three decimal places.
A)$3.570 billion
B)$24.990 billion
C)$19.763 billion
D)$29.645 billion
E)$12.495 billion

Accepted Answer

The answer of Suppose that  model the revenue (in...

Question 5

Find the area of the region bounded by the graphs of the algebraic functions. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

To find the area, we need to find the points of intersection of the graphs. 
$$\begin{cases} y = x^3 - 4x^2 + 5x \ y = x^2 - 3x + 2 \end{cases}$$
Setting them equal to each other, 
$$x^3 - 4x^2 + 5x = x^2 - 3x + 2$$
$$x^3 - 5x^2 + 8x - 2 = 0$$
We can use synthetic division or rational root theorem to find one root as 2. Dividing by $x - 2$ gives us
$$x^2 - 3x +1$$
Using quadratic formula, the other two roots are $\dfrac{3 \pm \sqrt{5}}{2}$
Now we can integrate to find the area:
$$\int_0^{\frac{3-\sqrt{5}}{2}} (x^2 - 3x +2) dx + \int_{\frac{3-\sqrt{5}}{2}}^2 (x^3 - 4x^2 +5x - x^2 +3x - 2) dx + \int_2^{\frac{3+\sqrt{5}}{2}} (x^2 - 3x +2) dx$$
Simplifying,
$$\int_0^{\frac{3-\sqrt{5}}{2}} (x^2 - 3x +2) dx + \int_{\frac{3-\sqrt{5}}{2}}^2 (x^3 - 5x^2 +2x - 2) dx + \int_2^{\frac{3+\sqrt{5}}{2}} (x^2 - 3x +2) dx$$
Evaluating the integrals gives us the final answer $\dfrac{8 - \sqrt{5}}{6} + \dfrac{7-2\sqrt{5}}{6} + \dfrac{8 - \sqrt{5}}{6} = \dfrac{17-4\sqrt{5}}{3}$, which is approximately 1.407. The region is bounded by the graphs of $y=2x+2$ and $y=-\frac{1}{2}x+\frac{7}{2}$.

Setting these two equations equal gives $2x+2=-\frac{1}{2}x+\frac{7}{2}$, which simplifies to $x=1$. Thus, we can find the $y$-values of the intersection by plugging in $x=1$: $y_{1}=2\cdot 1+2=4$ and $y_{2}=-\frac{1}{2}\cdot 1+\frac{7}{2}=\frac{5}{2}$.

The intersection point is $(1,\frac{5}{2})$.

Now we need to integrate the two functions with respect to $x$ to find the area:
\begin{align*}
A &= \int_{0}^{1} (2x+2) dx - \int_{0}^{1} \left(-\frac{1}{2}x+\frac{7}{2}ight) dx \
&= \left[x^{2}+2xight]_{0}^{1} - \left[-\frac{1}{4}x^{2}+\frac{7}{2}xight]_{0}^{1} \
&= \left(1+2ight)-\left(-\frac{1}{4}+\frac{7}{2}ight) \
&= \boxed{\frac{15}{4}}
\end{align*}

Question 6

Concrete sections for the new building have the dimensions (in meters) and shape as shown in the figure (the picture is not necessarily drawn to scale).Find the area of the face of the
Section superimposed on the rectangular coordinate system.Round your answer to three decimal
Places. 
A)25.031 
B)31.075 
C)29.151 
D)30.515 
E)28.031

Accepted Answer

The answer of Concrete sections for the new building have...

Question 7

Find the area of the region bounded by equations by integrating (i) with respect to  and (ii) with respect to y. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 8

Find the area of the region bounded by the graphs of the equations. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 9

Find the area of the region bounded by the graphs of the algebraic functions. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

To find the area bounded by the two curves, we need to find the points of intersection first. Setting the two equations equal to each other, we get:
$$25-x^2=x+4$$
$$x^2+x-29=0$$
Solving for $x$ gives $x=4$ (rejecting the negative solution).

Thus the area of the region bounded by the two curves is:

$$\int_{-4}^{4} (25-x^2)-(x+4) dx=\int_{-4}^{4} 21-x-x^2 dx$$

Using the power rule, the integral evaluates to:

$$\left[ 21x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-4}^{4}= \frac{512}{3}$$

So the answer is C.

Question 10

Set up the definite integral that gives the area of the region bounded by the graph of 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Set up the definite integral that gives...

Question 11

Find the area of the region bounded by the equations by integrating (i) with respect to x and
(ii) with respect to y. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 12

The chief financial officer of a company reports that profits for the past fiscal year were  million.The officer predicts that profits for the next 7 years will grow at a continuous
Annual rate somewhere between  and 6%.Estimate the cumulative difference in total profit over
The 7 years based on the predicted range of growth rates.Round your answer to three decimal places.
A)$445.736 billion
B)$30.221 billion
C)$7.023 billion
D)$18.710 billion
E)$57.880 billion

Accepted Answer

The answer of The chief financial officer of a company...

Question 13

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by  in the first quadrant about the -axis.
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Set up and evaluate the integral that...

Question 14

The surface of a machine part is the region between the graphs of  and  as shown in the figure.Find k if the parabola is tangent to the graph of  .Round your answer to three decimal places. 
A)3.125
B)0.160
C)0.080
D)0.320
E)6.250

Accepted Answer

The answer of The surface of a machine part is...

Question 15

Find the area of the region bounded by the graphs of the function  Round your answer to three decimal places.
A)20.723
B)11.182
C)6.238
D)10.362
E)22.364

Accepted Answer

The answer of Find the area of the region bounded...

Question 16

If the accumulation function 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of If the accumulation function 
A) 
B) 
C)...

Question 17

Find the area of the region bounded by the graphs of the algebraic functions. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 18

Find the area of the region bounded by the graphs of the equations. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 19

Find the area of the region bounded by the graphs of the algebraic functions. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Find the area of the region bounded...

Question 20

The surface of a machine part is the region between the graphs of  and  as shown in the figure.Find the area of the surface of the machine part.Round your answer to five decimal places. 
A)1.66667
B)0.41667
C)0.01333
D)33.33333
E)8.33333

Accepted Answer

To find the surface area, we need to integrate the function (f(x) - g(x)) over the interval [0,2]. 
f(x) = 1 + cos(πx) + 0.5cos(2πx) 
g(x) = cos(πx) 
So, the surface area = ∫[0,2] (f(x) - g(x)) dx. 
= ∫[0,2] (1 + 0.5cos(2πx)) dx. 
= [x + 0.25(1/π)sin(2πx)] [from 0 to 2] 
= 2 + 0.25(1/π)sin(4π) - 0.25(1/π)sin(0) 
= 2 + 0.25(1/π)sin(4π) 
= 2 + 0.25(1/π)*0 
= 2 
Rounding to five decimal places, we get 2.00000, which is closest to 8.33333 (option E).

Quiz 7: Applications of Integration

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by axis.

Concrete sections for the new building have the dimensions (in meters) and shape as shown in the figure (the picture is not necessarily drawn to scale) .One cubic meter of concrete Weighs 4320 pounds.Find the weight of the section.Round your answer to the nearest pound.

Find the area of the region bounded by the graphs of the function Round your answer to three decimal places.

Find the area of the region bounded by the graphs of the algebraic functions.

Concrete sections for the new building have the dimensions (in meters) and shape as shown in the figure (the picture is not necessarily drawn to scale) .Find the area of the face of the Section superimposed on the rectangular coordinate system.Round your answer to three decimal Places.

Find the area of the region bounded by equations by integrating (i) with respect to and (ii) with respect to y.

Find the area of the region bounded by the graphs of the equations.

Find the area of the region bounded by the graphs of the algebraic functions.

Set up the definite integral that gives the area of the region bounded by the graph of

Find the area of the region bounded by the equations by integrating (i) with respect to x and (ii) with respect to y.

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by in the first quadrant about the -axis.

The surface of a machine part is the region between the graphs of and as shown in the figure.Find k if the parabola is tangent to the graph of .Round your answer to three decimal places.

Find the area of the region bounded by the graphs of the function Round your answer to three decimal places.

If the accumulation function

Find the area of the region bounded by the graphs of the algebraic functions.

Find the area of the region bounded by the graphs of the equations.

Find the area of the region bounded by the graphs of the algebraic functions.

The surface of a machine part is the region between the graphs of and as shown in the figure.Find the area of the surface of the machine part.Round your answer to five decimal places.