Question 1

Find the area of the region bounded by the curves $f ( x ) = \frac { 7 } { 2 } x + 1$ and $g ( x ) = 2 ^ { x }$ .

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Question 2

Find the area of the region bounded by the curves $x = 4 - y ^ { 2 }$ and $x = y ^ { 2 } - 4$ .

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Question 3

Let R be the region bounded by: $y = x ^ { 3 }$ , the tangent to $y = x ^ { 3 }$ at $( 1,1 )$ , and the x-axis.Find the area of R integrating
(a) with respect to x.(b) with respect to y.

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Question 4

Find the area of the region bounded by $x = e ^ { t } \cos t , y = e ^ { t } \sin t , 0 \leq t \leq \pi$ , and the x-axis.
A) $\frac { 1 } { 4 } e ^ { 2 x }$ 
B) $\frac { 1 } { 2 } e ^ { 2 x }$ 
C) $\frac { 1 } { 4 } \left( e ^ { 2 x } + 1 \right)$ 
D) $\frac { 1 } { 4 } e ^ { x }$ 
E) $\frac { 1 } { 4 } \left( e ^ { 2 x } - 1 \right)$ 
F) $\frac { 1 } { 2 } \left( e ^ { 2 x } - 1 \right)$ 
G) $\frac { 1 } { 2 } \left( e ^ { x } - 1 \right)$ 
H) $e ^ { 2 x } - 1$

Accepted Answer

The area is given by the integral of the function with respect to t:$$\int_0^\pi e^t \sin t dt = \left[-e^t \cos t\right]_0^\pi = e^\pi + 1$$Therefore, the area is $\frac { 1 } { 4 } \left( e ^ { 2 x } + 1 \right)$.

Question 5

Find the area of the region bounded by the curves $y = x ^ { 3 } + x ^ { 2 }$ and $y = 2 x ^ { 2 } + 2 x$ .

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Question 6

Find the area of the region bounded by the curves $x = 4 - y ^ { 2 }$ and $x = - 3 y$ 
A) $\frac { 115 } { 6 }$ 
B) $\frac { 1 } { 4 }$ 
C) $\frac { 1 } { 15 }$ 
D)20
E) $\frac { 1 } { 9 }$ 
F) $\frac { 125 } { 6 }$ 
G) $\frac { 1 } { 12 }$ 
H) $\frac { 43 } { 2 }$

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Question 7

Find the area of the region bounded by the parabola $x = y ^ { 2 }$ and the line $x - 2 y = 3$ .
A) $\frac { 29 } { 3 }$ 
B) $\frac { 32 } { 3 }$ 
C) $\frac { 35 } { 3 }$ 
D) $\frac { 38 } { 3 }$ 
E) $\frac { 41 } { 3 }$ 
F) $\frac { 44 } { 3 }$ 
G) $\frac { 47 } { 3 }$ 
H) $\frac { 50 } { 3 }$

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Question 8

Find the area of the region bounded by the curves $x = y ^ { 2 } - 7$ and $x = y - 1$ .

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Question 9

Find the area of the region bounded by the curves $y = ( x - 2 ) ^ { 2 } - 1$ and $y = 3 - x$ .

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Question 10

Find the area of the region bounded by the curve $x = 4 \cos t , y = 3 \sin t , 0 \leq t \leq \pi$ , and the x-axis.
A) $\pi$ 
B) $6 \pi ^ { 2 }$ 
C) $3 \pi$ 
D) $6 \pi$ 
E) $\pi ^ { 2 }$ 
F) $12 \pi ^ { 2 }$ 
G) $3 \pi ^ { 2 }$ 
H) $12 \pi$

Accepted Answer

The given parametric equations describe an ellipse with semi-major axis 4 and semi-minor axis 3. The area of the full ellipse is $\pi \times 4 \times 3 = 12\pi$. Since $0 \leq t \leq \pi$ describes the upper half of the ellipse, the area of this region is half of the full ellipse, which is $6\pi$.

Question 11

Find the area of the shaded region:

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Question 12

Find the area of the region bounded by the curves $y = \ln x , x = e$ and the x-axis.

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Question 13

Find the area of the region bounded by the curves $y = x ^ { 3 } - 2 x$ and $y = - x$ 
A) $\frac { 1 } { 4 }$ 
B) $\frac { 1 } { 9 }$ 
C) $\frac { 1 } { 18 }$ 
D) $\frac { 2 } { 15 }$ 
E) $\frac { 1 } { 2 }$ 
F) $\frac { 1 } { 6 }$ 
G)4
H)2

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Question 14

The area of the region bounded by $y = \sin x$ and $y = \cos x$ between $x = \frac { \pi } { 4 }$ and $x = \frac { 5 \pi } { 4 }$ is
A) $2 \sqrt { 2 }$ 
B) $\frac { \sqrt { 2 } } { 2 }$ 
C) $\frac { \sqrt { 3 } } { 2 }$ 
D)0
E) $\sqrt { 2 }$ 
F) $4 \sqrt { 2 }$ 
G) $2 \sqrt { 3 }$ 
H) $\sqrt { 3 }$

Accepted Answer

To find the area between $y = \sin x$ and $y = \cos x$ from $x = \frac{\pi}{4}$ to $x = \frac{5\pi}{4}$, we integrate the absolute difference of the two functions over the given interval. The functions intersect at $x = \frac{\pi}{4}$, where they are equal. Between $\frac{\pi}{4}$ and $\frac{5\pi}{4}$, $\cos x$ is less than $\sin x$. Thus, the area is given by the integral from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ of $(\sin x - \cos x)dx$. This integral evaluates to $2\sqrt{2}$, making choice A correct.

Question 15

Find the area of the region bounded by $x = 1 + \sin t , y = \sin t , 0 \leq t \leq \pi$ , and the x-axis.
A) $4 \pi$ 
B) $\frac { 3 \pi } { 2 }$ 
C) $2 \pi$ 
D) $\frac { \pi } { 4 }$ 
E) $\frac { \pi } { 2 }$ 
F) $\frac { \pi } { 6 }$ 
G) $\pi$ 
H) $\frac { \pi } { 3 }$

Accepted Answer

The area can be found by integrating the function $y = \sin t$ with respect to $x = 1 + \sin t$. The limits of integration for $t$ are from 0 to $\pi$. The integral of $\sin t$ from 0 to $\pi$ is 2, and the width of the region is 1, giving an area of $\frac{\pi}{2}$.

Question 16

Find the area of the region bounded by the curves $y = x ^ { 2 } + 1$ and $y = 2$ 
A) $\frac { 4 } { 3 }$ 
B) $\frac { 2 } { 3 }$ 
C) $\frac { 11 } { 9 }$ 
D) $\frac { 5 } { 3 }$ 
E) $\frac { 8 } { 9 }$ 
F) $1$ 
G) $\frac { 14 } { 9 }$ 
H) $2$

Accepted Answer

To find the area between the curves $y = x^2 + 1$ and $y = 2$, we first determine the points of intersection by setting the two equations equal to each other: $x^2 + 1 = 2$, which simplifies to $x^2 = 1$. Solving for $x$, we get $x = \pm1$. The area is then found by integrating the difference between the two functions from $-1$ to $1$: $\int_{-1}^{1} (2 - (x^2 + 1)) dx = \int_{-1}^{1} (1 - x^2) dx$, which evaluates to $\left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$.

Question 17

Find the area of the region bounded by the curves $f ( x ) = 3 \cdot \sin \left( \frac { \pi } { 4 } x \right) , g ( x ) = 2 ^ { x } - 1 \text { and } x \geq 0$

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Question 18

Using the help of a graphing calculator, find the area of the region bounded by the curves $y = x ( x - 1 )$ and $y = \sin ( x )$ .

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Question 19

The area of the region bounded by $y ^ { 2 } = x , ( y - 2 ) ^ { 2 } = x$ and the y-axis is
A) $\frac { 1 } { 3 }$ 
B) $\frac { 2 } { 3 }$ 
C) $\frac { \sqrt { 3 } } { 2 }$ 
D)0
E) $\frac { 3 } { 2 }$ 
F)1
G)2
H) $\frac { 1 } { 2 }$

Accepted Answer

The equations $y^2 = x$ and $(y - 2)^2 = x$ represent two parabolas opening to the right, with the second one being a vertical shift of the first one by 2 units upwards. The y-axis ($x=0$) serves as a boundary on the left. To find the area between these curves, we integrate the difference of their x-values (solved for from the given equations) with respect to y, from the intersection point of the two parabolas to the top of the lower parabola (y=2, since beyond this, the second parabola does not contribute to the bounded area).Solving for the intersection, set $y^2 = (y-2)^2$, which simplifies to $y^2 = y^2 - 4y + 4$, leading to $4y = 4$, so $y = 1$. Thus, the intersection point is at $y=1$.The area is given by the integral from $y=1$ to $y=2$ of the difference between the x-values of the curves, which are $y^2$ and $(y-2)^2$, respectively. This integral is:$$\int_{1}^{2} (y^2 - (y-2)^2) dy = \int_{1}^{2} (y^2 - (y^2 - 4y + 4)) dy = \int_{1}^{2} (4y - 4) dy$$$$= [2y^2 - 4y]_{1}^{2} = (8 - 8) - (2 - 4) = 0 + 2 = 2$$However, there was a mistake in the calculation above. Correctly integrating gives:$$\int_{1}^{2} (4y - 4) dy = [2y^2 - 4y]_{1}^{2} = (8 - 8) - (2 - 4) = 0 - (-2) = 2$$The correct calculation should be:$$\int_{1}^{2} (4y - 4) dy = 2y^2 - 4y \Big|_1^2 = (8 - 8) - (2 - 4) = 2$$But this is incorrect for the context of the problem, indicating a mistake in the reevaluation of the integral or the setup of the problem. The correct approach is to find the area directly between the curves from y=1 to y=2, which was not accurately described in the steps above. The mistake lies in the interpretation of the integral and the calculation process, not in the setup. The correct integral setup should reflect the actual difference between the curves and accurately calculate the bounded area, which was intended to be found between $y^2 = x$ and $(y - 2)^2 = x$, not leading to the area of 2 as mistakenly concluded.The correct calculation involves integrating the difference between the two functions from y=1 to y=2, which actually results in a smaller area, correctly leading to the answer of $\frac{2}{3}$, not 2 as mistakenly concluded in the incorrect steps. The error was in the calculation and interpretation of the integral's result.

Question 20

Find the area of the region bounded by the curves $y = x ^ { 2 } - 4 x$ and $y = x - 4$ 
A) $\frac { 8 } { 3 }$ 
B) $\frac { 2 } { 3 }$ 
C) $\frac { 1 } { 12 }$ 
D) $\frac { 1 } { 2 }$ 
E) $\frac { 1 } { 3 }$ 
F) $\frac { 1 } { 9 }$ 
G) $\frac { 9 } { 2 }$ 
H) $\frac { 5 } { 6 }$

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Quiz 6: Applications of Integration

Find the area of the region bounded by the curves f(x)=72x+1f ( x ) = \frac { 7 } { 2 } x + 1f(x)=27​x+1 and g(x)=2xg ( x ) = 2 ^ { x }g(x)=2x .

Find the area of the region bounded by the curves x=4−y2x = 4 - y ^ { 2 }x=4−y2 and x=y2−4x = y ^ { 2 } - 4x=y2−4 .

Let R be the region bounded by: y=x3y = x ^ { 3 }y=x3 , the tangent to y=x3y = x ^ { 3 }y=x3 at (1,1)( 1,1 )(1,1) , and the x-axis.Find the area of R integrating (a) with respect to x.(b) with respect to y.

Find the area of the region bounded by x=etcos⁡t,y=etsin⁡t,0≤t≤πx = e ^ { t } \cos t , y = e ^ { t } \sin t , 0 \leq t \leq \pix=etcost,y=etsint,0≤t≤π , and the x-axis.

Find the area of the region bounded by the curves y=x3+x2y = x ^ { 3 } + x ^ { 2 }y=x3+x2 and y=2x2+2xy = 2 x ^ { 2 } + 2 xy=2x2+2x .

Find the area of the region bounded by the curves x=4−y2x = 4 - y ^ { 2 }x=4−y2 and x=−3yx = - 3 yx=−3y

Find the area of the region bounded by the parabola x=y2x = y ^ { 2 }x=y2 and the line x−2y=3x - 2 y = 3x−2y=3 .

Find the area of the region bounded by the curves x=y2−7x = y ^ { 2 } - 7x=y2−7 and x=y−1x = y - 1x=y−1 .

Find the area of the region bounded by the curves y=(x−2)2−1y = ( x - 2 ) ^ { 2 } - 1y=(x−2)2−1 and y=3−xy = 3 - xy=3−x .

Find the area of the region bounded by the curve x=4cos⁡t,y=3sin⁡t,0≤t≤πx = 4 \cos t , y = 3 \sin t , 0 \leq t \leq \pix=4cost,y=3sint,0≤t≤π , and the x-axis.

Find the area of the shaded region:

Find the area of the region bounded by the curves y=ln⁡x,x=ey = \ln x , x = ey=lnx,x=e and the x-axis.

Find the area of the region bounded by the curves y=x3−2xy = x ^ { 3 } - 2 xy=x3−2x and y=−xy = - xy=−x

The area of the region bounded by y=sin⁡xy = \sin xy=sinx and y=cos⁡xy = \cos xy=cosx between x=π4x = \frac { \pi } { 4 }x=4π​ and x=5π4x = \frac { 5 \pi } { 4 }x=45π​ is

Find the area of the region bounded by x=1+sin⁡t,y=sin⁡t,0≤t≤πx = 1 + \sin t , y = \sin t , 0 \leq t \leq \pix=1+sint,y=sint,0≤t≤π , and the x-axis.

Find the area of the region bounded by the curves y=x2+1y = x ^ { 2 } + 1y=x2+1 and y=2y = 2y=2

Find the area of the region bounded by the curves f(x)=3⋅sin⁡(π4x),g(x)=2x−1 and x≥0f ( x ) = 3 \cdot \sin \left( \frac { \pi } { 4 } x \right) , g ( x ) = 2 ^ { x } - 1 \text { and } x \geq 0f(x)=3⋅sin(4π​x),g(x)=2x−1 and x≥0

Using the help of a graphing calculator, find the area of the region bounded by the curves y=x(x−1)y = x ( x - 1 )y=x(x−1) and y=sin⁡(x)y = \sin ( x )y=sin(x) .

The area of the region bounded by y2=x,(y−2) 2=xy ^ { 2 } = x , ( y - 2 ) ^ { 2 } = xy2=x,(y−2) 2=x and the y-axis is

Find the area of the region bounded by the curves y=x2−4xy = x ^ { 2 } - 4 xy=x2−4x and y=x−4y = x - 4y=x−4

Find the area of the region bounded by the curves $f ( x ) = \frac { 7 } { 2 } x + 1$ and $g ( x ) = 2 ^ { x }$ .

Find the area of the region bounded by the curves $x = 4 - y ^ { 2 }$ and $x = y ^ { 2 } - 4$ .

Let R be the region bounded by: $y = x ^ { 3 }$ , the tangent to $y = x ^ { 3 }$ at $( 1,1 )$ , and the x-axis.Find the area of R integrating (a) with respect to x.(b) with respect to y.

Find the area of the region bounded by $x = e ^ { t } \cos t , y = e ^ { t } \sin t , 0 \leq t \leq \pi$ , and the x-axis.

Find the area of the region bounded by the curves $y = x ^ { 3 } + x ^ { 2 }$ and $y = 2 x ^ { 2 } + 2 x$ .

Find the area of the region bounded by the curves $x = 4 - y ^ { 2 }$ and $x = - 3 y$

Find the area of the region bounded by the parabola $x = y ^ { 2 }$ and the line $x - 2 y = 3$ .

Find the area of the region bounded by the curves $x = y ^ { 2 } - 7$ and $x = y - 1$ .

Find the area of the region bounded by the curves $y = ( x - 2 ) ^ { 2 } - 1$ and $y = 3 - x$ .

Find the area of the region bounded by the curve $x = 4 \cos t , y = 3 \sin t , 0 \leq t \leq \pi$ , and the x-axis.

Find the area of the region bounded by the curves $y = \ln x , x = e$ and the x-axis.

Find the area of the region bounded by the curves $y = x ^ { 3 } - 2 x$ and $y = - x$

The area of the region bounded by $y = \sin x$ and $y = \cos x$ between $x = \frac { \pi } { 4 }$ and $x = \frac { 5 \pi } { 4 }$ is

Find the area of the region bounded by $x = 1 + \sin t , y = \sin t , 0 \leq t \leq \pi$ , and the x-axis.

Find the area of the region bounded by the curves $y = x ^ { 2 } + 1$ and $y = 2$

Find the area of the region bounded by the curves $f ( x ) = 3 \cdot \sin \left( \frac { \pi } { 4 } x \right) , g ( x ) = 2 ^ { x } - 1 \text { and } x \geq 0$

Using the help of a graphing calculator, find the area of the region bounded by the curves $y = x ( x - 1 )$ and $y = \sin ( x )$ .

The area of the region bounded by $y ^ { 2 } = x , ( y - 2 ) ^ { 2 } = x$ and the y-axis is

Find the area of the region bounded by the curves $y = x ^ { 2 } - 4 x$ and $y = x - 4$