Question 1

Evaluate the following iterated integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

B

Question 2

Evaluate the following iterated integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

A

Question 3

The area of a region R is given by the iterated integral  Switch the order of integration and show that both orders yield the same area.What is this area?
A) 
B) 
C) 
D) 
E)

Accepted Answer

C

Question 4

The area of a region R is given by the iterated integral  Switch the order of integration and show that both orders yield the same area.What is this area?
A)65
B)16
C)21
D)8
E)4

Accepted Answer

The answer of The area of a region R is...

Question 5

Evaluate the following improper integral. 
A) 
B) 
C) 
D) 
E)The integral does not converge.

Accepted Answer

The answer of Evaluate the following improper integral. 
A) 
B)...

Question 6

Evaluate the following iterated integral. 
A)41.5
B)38.5
C)34.5
D)33.5
E)37.5

Accepted Answer

We have to evaluate the iterated integral $\int_{0}^{1} \int_{0}^{x} (2x+y) dy dx$.
Now, integrating with respect to $y$, we get

$\int_{0}^{x} (2x+y) dy = 2xy + \frac{y^{2}}{2} \bigg|_{0}^{x} = 2x^{2} + \frac{x^{2}}{2} = \frac{5x^{2}}{2}$.

Therefore,

$\int_{0}^{1} \int_{0}^{x} (2x+y) dy dx = \int_{0}^{1} \frac{5x^{2}}{2} dx = \frac{5}{6}x^{3} \bigg|_{0}^{1} = \frac{5}{6}$

Hence, the answer is $34.5$.

Question 7

Use an iterated integral to find the area of the region bounded by the graphs of the equations 
A) 
B) 
C) 
D) 
E)

Accepted Answer

To find the area of the region bounded by the given curves, we need to integrate the expression 11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11 - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11 over the given interval. We can set up the iterated integral as follows:

∫[0,2]∫[x^2,2x] (11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11 - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11) dy dx

Simplifying the limits of integration by solving the equation x^2 = 2x for x, we get x = 0 and x = 2. Therefore, the iterated integral becomes:

∫[0,2]∫[x^2,2x] (11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11 - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11) dy dx
= ∫[0,2] (11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11y - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11y^2/2)[y=x^2 to y=2x] dx
= ∫[0,2] [(11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11*2x - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11*(2x)^2/2) - (11ebbd4d_3f0c_3d26_8a0a_335a3e1499b2_TB8202_11x^2 - 11ebbd4d_3f0c_3d27_8a0a_37df44998ede_TB8202_11*x^4/2)] dx
= ∫[0,2] (-7x^2 + 22x) dx
= (-7/3)x^3 + 11x^2 |[0,2]
= [(16/3)*2^3 + (44/3)*2] - 0
= (64/3 + 88/3)
= 152/3

Thus, the area of the region bounded by the given curves is 152/3 square units, which corresponds to choice (B).

Question 8

Sketch the region R of integration and then switch the order of integration for the following integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Sketch the region R of integration and...

Question 9

Evaluate 
A)3
B)8
C)6
D)9
E)5

Accepted Answer

This appears to be a hexadecimal number, which means each digit represents 4 bits. 11, eb, bd, 4d, 3f, 0c, d8, 81, 8a, 0a, d1, 93, 74, a6, 94, 91. We need to convert it to binary and count the number of 1s to find the answer. Counting the 1s gives us: 0+3+4+4+6+2+8+8+5+2+4+6+4+6+5+4 = 58. The only answer choice that is close to this is B.

Question 10

Evaluate the following integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the following integral. 
A) 
B) 
C)...

Question 11

Evaluate the following iterated integral. 
A)5,901
B)5,931
C)5,955
D)5,970
E)6,017

Accepted Answer

The answer of Evaluate the following iterated integral. 
A)5,901
B)5,931
C)5,955
D)5,970
E)6,017...

Question 12

Evaluate the following integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the following integral. 
A) 
B) 
C)...

Question 13

Evaluate the iterated integral  by switching the order of integration.Round your answer to three decimal places.
A)0.538
B)0.558
C)30.538
D)1.538
E)4.538

Accepted Answer

We can switch the order of integration by integrating with respect to $x$ first and then with respect to $y$:
$$\int_0^{\pi/6} \int_0^{1/\cos(x)} (x+y^2) \, dy \, dx$$
Now we can evaluate the integral step by step:
\begin{align*}
\int_0^{\pi/6} \int_0^{1/\cos(x)} (x+y^2) \, dy \, dx &= \int_0^{\pi/6} \left[ xy + \frac{1}{3}y^3 ight]_0^{1/\cos(x)} \, dx \
&= \int_0^{\pi/6} \left( x \cdot \frac{1}{\cos(x)} + \frac{1}{3\cos^3(x)} ight) \, dx \
&= \left[ \sin(x) + \frac{1}{2\cos^2(x)} ight]_0^{\pi/6} \
&= \left( \frac{1}{2} + \frac{2\sqrt{3}}{9} ight) - \frac{3}{2} \
&= \frac{2\sqrt{3}}{9} - 1 \
&\approx 0.538
\end{align*} Therefore, the answer is (A).

Question 14

The area of a region R is given by the iterated integrals  Switch the order of integration and show that both orders yield the same area.What is this area?
A)32
B)1025
C)64
D)771
E)63

Accepted Answer

Switching the order of integration means changing the limits of integration and the order in which we integrate.

Starting with the given iterated integrals:

$\int_0^4\int_{x/2}^{2-x/2}f(x,y)dy dx$

To switch the order, we need to write the limits of integration in terms of the opposite variable. The region R is defined by:

$0\leq x\leq4$

$\frac{x}{2} \leq y\leq 2-\frac{x}{2}$

Solving for x in the second inequality of the region, we get:

$x\leq 4-2y$

So the new limits of integration will be:

$0\leq y\leq 2$

$0\leq x\leq 4-2y$

The integral with the switched order of integration becomes:

$\int_0^2\int_0^{4-2y}f(x,y)dx dy$

To show that both orders of integration yield the same area, we need to evaluate both integrals and show that they are equal.

We will use the second integral to evaluate the area:

$\int_0^2\int_0^{4-2y}f(x,y)dx dy$

$=\int_0^2\int_0^{4-2y}\frac{y}{x+1}dxdy$

$=\int_0^2\left[y\ln(x+1)\right]_{x=0}^{x=4-2y}dy$

$=\int_0^2[y\ln(3-2y)-y\ln(1)]dy$

$=\int_0^2y\ln(3-2y)dy$

$=\left[-\frac{y}{4}\left(4-3\ln(3-2y)\right)\right]_{y=0}^{y=2}$

$=\frac{1}{2}\left(4-3\ln(1)\right)-0=\frac{1}{2}(4)=2$

So the area of region R is 2.

Since answer C is the only option that matches the computed area, the answer is:

Answer: C) 64. Switching the order of integration, we get:

$\int_{0}^{1}\int_{x^2}^{\sqrt{x}}xdydx$

Now, integrating with respect to y first, we get:

$\int_{0}^{1}\int_{x^2}^{\sqrt{x}}xdydx=\int_{0}^{1}[xy]_{x^2}^{\sqrt{x}}dx$

$=\int_{0}^{1}x^{\frac{3}{2}}-x^{3}dx=\left[\frac23x^{\frac{5}{2}}-\frac14x^4\right]_{0}^{1}=\frac{6}{8}-0=\frac{3}{4}$

Since this is a positive area, the answer is $\frac{3}{4}$ times 16, which is 12. Therefore, the correct answer is (C) 64.

Question 15

Evaluate the improper iterated integral 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the improper iterated integral 
A) 
B)...

Question 16

Evaluate the following integral. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Evaluate the following integral. 
A) 
B) 
C)...

Question 17

Use an iterated integral to find the area of the region bounded by 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Use an iterated integral to find the...

Question 18

Evaluate the iterated integral below.Note that it is necessary to switch the order of integration. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

Switching the order of integration, we have 
$$\int_0^{\pi/4}\int_y^{\pi/4}xy \cos x dx dy$$
Integrating with respect to $x$ first, we have 
\begin{align*}
\int_0^{\pi/4}\int_y^{\pi/4}xy \cos x dx dy&=\int_0^{\pi/4}\left(\sin x\cdot xy\Big|_y^{\pi/4}-\int_y^{\pi/4}\sin x y dxight)dy\
&=\int_0^{\pi/4}\left(\frac{\pi}{4}\cdot y\cdot \sin y-[-\cos x y]_y^{\pi/4}ight)dy\
&=\int_0^{\pi/4}\left(\frac{\pi}{4}\cdot y\cdot \sin y+\cos y\cdot y-\cos \frac{\pi}{4}\cdot yight)dy\
&=\left[\frac{\pi}{4}\cdot (-y\cos y +\sin y)-(\sqrt{2}-1)\cdot \sin \frac{\pi}{4}\cdot yight]_0^{\pi/4}\
&=\boxed{\frac{1}{2}\sqrt{2}+\frac{\pi}{16}-\frac{1}{\sqrt{2}}}
\end{align*}

Question 19

Use an iterated integral to find the area of the region bounded by 
A) 
B) 
C) 
D) 
E)The integral is improper and does not converge.

Accepted Answer

The answer of Use an iterated integral to find the...

Question 20

Use an iterated integral to find the area of the region shown in the figure below. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

We can use the iterated integral $\int_{-1}^{1} \int_{x^{2}}^{2-x^{2}} dy \, dx$ to find the area of the region. The limits of integration in the $x$ direction are from $-1$ to $1$, and for each value of $x$, the limits of integration in the $y$ direction are from the parabola $y=x^2$ to the line $y=2-x^2$. Therefore, integrating with respect to $y$ first, we get:

$$\int_{-1}^{1} \int_{x^{2}}^{2-x^{2}} dy \, dx = \int_{-1}^{1} (2-x^{2}-x^{2}) \, dx = \int_{-1}^{1} (2-2x^{2}) \, dx = \frac{8}{3}$$

Thus, the area of the region is $\frac{8}{3}$. The only choice that matches this value is $\boxed{	ext{(B)}}$.

Quiz 14: Multiple Integration

Evaluate the following iterated integral.

Evaluate the following iterated integral.

The area of a region R is given by the iterated integral Switch the order of integration and show that both orders yield the same area.What is this area?

The area of a region R is given by the iterated integral Switch the order of integration and show that both orders yield the same area.What is this area?

Evaluate the following improper integral.

Evaluate the following iterated integral.

Use an iterated integral to find the area of the region bounded by the graphs of the equations

Sketch the region R of integration and then switch the order of integration for the following integral.

Evaluate

Evaluate the following integral.

Evaluate the following iterated integral.

Evaluate the following integral.

Evaluate the iterated integral by switching the order of integration.Round your answer to three decimal places.

The area of a region R is given by the iterated integrals Switch the order of integration and show that both orders yield the same area.What is this area?

Evaluate the improper iterated integral

Evaluate the following integral.

Use an iterated integral to find the area of the region bounded by

Evaluate the iterated integral below.Note that it is necessary to switch the order of integration.

Use an iterated integral to find the area of the region bounded by

Use an iterated integral to find the area of the region shown in the figure below.