Integrals Test

Result

Integrals Test - 6

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Weekly Quiz Competition

Question 1
1 / -0.25

In the definite integral , the variable of integration is called

A
Dependent variable

B
Dummy variable

C
Constant

D
Constant of integration
Question 2
1 / -0.25

Express the shaded area in the form of an integral.

A

B

C

D
None of the above.

Solution

As the curve goes from c to d and the equation is x = f(y)
So the shaded area is ∫(c to d)f(y)dy
Question 3
1 / -0.25

Evaluate as limit of sum

A
20/5

B
15/2

C
20/3

D
3/20

Solution

∫(0 to 2)(x² + x + 1)dx
= (0 to 2) [x³ /3 + x² /2 + x]^½
= [8/3 + 4/2 + 2]
= 40/6
= 20/3
Question 4
1 / -0.25

Evaluate as limit of sum

A
2

B
-2/3

C
4

D
1

Solution
Question 5
1 / -0.25

The value of definite integral depends on

A
The function, the interval and the variable of integration

B
The function and the variable of integration

C
The function and the interval

D
The interval and the variable of integration
Question 6
1 / -0.25

Find

A
π/4

B
π/2

C
4 π

D
2 π

Solution

Using trigonometric identities, we have
cos² x=cos² x-sin² x -(1) and cos² x+sin² x =1 -(2)
cos² x=1-sin² x , substituting this in equation (1) we get
cos² x=1-sin² x-sin² x=1-2sin² x
So,cos² x=1-2sin² x
2sin² x=1-cos² x
Question 7
1 / -0.25

Evaluate as limit of sum

A
20

B
2

C
12

D
2/3
Question 8
1 / -0.25

The value of is:

A
0

B
-1

C
1

D
2
Question 9
1 / -0.25

Evaluate as limit of sum

A
12 sq units

B
24 sq units

C
48 sq units

D
None of the above

Solution

∫(0 to 4)3x dx
= [3x² /2] (0 to 4)
[3(4)² ] / 2
= 24 sq unit
Question 10
1 / -0.25

The value of is:

A
π/4

B
π/2

C
1

D
π/8

Solution

∫(0 to 3)1/[(3)² - (x)^2]^½
∫1/[(a)² - (x)² ] = sin^-1 (x/a)
= [sin^-1 (x/3)](0 to 3)
= sin^-1 [3/3] - sin^-1 [0/3]
= sin^-1 [1]
= π/2