Applications of the Integrals Test

Result

Applications of the Integrals Test - 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

The area between the curves y = x² and y = x³ is:

A
1/12 sq.units

B
1/8 sq.units

C
1/10 sq.unit

D
1/6 sq.units

Solution

y = x² and y = x³
To find point of intersections :
x² = x³
x² (x-1) = 0
So, x = 0,1
POI are (0,0) &(1,1)

= [x³ /3 –x⁴ /4]_x=1 - [x³ /3 –x⁴ /4]_x=0
= (1/3 –1/4) –(0)
= 1/12 sq units
Question 2
1 / -0.25

The area bounded by the curve y² = 16x , x = 1, x = 3 and X-axis is:

A

B

C

D

Solution
Question 3
1 / -0.25

The shaded area is equal to:

A
29/3 unit²

B
31/3 unit²

C
32/3 unit²

D
28/3 unit²

Solution
Question 4
1 / -0.25

The curves x² + y² =16 and y² = 6x intersects at

A
(0, 2 √3)

B
(0, 2)

C
(2, 2 √3)

D
(2, 0)

Solution

x² + y² =16 and y² = 6x
So, to figure out the P.O.I
x² + 6x = 16
x² + 6x –16 = 0
(x+8) (x-2) = 0
x = -8, 2
For x = -8
y = (6*(-8))^1/2
= (-48)^1/2
No real value for y
For x = 2
y = (6*2)^1/2
= (12)^1/2
= 2*(3)^1/2
So, P.O.I is (2 , 2*(3)^1/2 )
Question 5
1 / -0.25

Find the area of region bounded by curve

A
3 π/2 sq. units

B
3 πsq. units

C
6 πsq. units

D
12 πsq. units

Solution
Question 6
1 / -0.25

The area of the region bounded by y = x² –2x and y = 4 –x² is.

A
7 sq. units

B
8 sq. units

C
9 sq. units

D
10 sq. units

Solution

To find area between y = x² –2x and y = 4 –x² ,
We need to find POI.
x² –2x = 4 –x²
2x² - 2x –4 = 0
x² –x –2 = 0
(x-2)(x+1) = 0
x = -1, 2

= [(-2x³ )/3 + x² + 4x]_x=2 - [(-2x³ )/3 + x² + 4x]_{x= -1}
= -16/3 + 4 + 8 –(2/3 + 1 - 4)
= 9 sq units
Question 7
1 / -0.25

The shaded area enclosed between the parabolas with equations y = 1 + 10x –2x² and y = 1 + 5x –x² is equal to:

A
21 sq. units

B
20 sq. units

C

D

Solution
Question 8
1 / -0.25

The area of the smaller region lying above the x-axis and included between the circle x² + y² = 2x and the parabola y² = x.

A

B

C

D

Solution

= π/4 –2/3
Question 9
1 / -0.25

The area bounded by the curves f(x) = x² + 1 and g(x) = x –1 on the interval [1,3] is:

A
26/3 units²

B
23/3 units²

C
22/3 units²

D
25/3 units²

Solution
Question 10
1 / -0.25

The area shaded in the given figure can be calculated by which of the following definite integral?

A

B

C

D

Solution