Indefinite Integrals MCQ Quiz - Objective Question with Answer for Indefinite Integrals - Download Free PDF

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

Let,

By solving through integration by parts, we get

where C is constant

Explanation:

To solve this, we can use

Let   so that 

dv = dx , so that v = x

Now, using integration by parts 

=

= 

So, the correct answer is 

The correct option is option 3.

Explanation:

= 

= 

= 

= 2(- cos (x/2) + sin (x/2))

= 2(sin (x/2) - cos (x/2))

(3) is true.

Substitude e^-x = t

-e^-x dx = dt

Explanation:

Put sin^-1 x = t

Concept:

Since the inside limit is in terms of x, therefore we have to integrate first the 'y' terms and convert whole expression in terms of x.

Calculation:

Given:

The Given Question is Wrong, Marks are allotted to all.

Explanation-

Given that, 

......

Expanding the options given,

option 1-

option 2-

option 3-

option 4-

So none of the options are matching with the correct answer. 

Explanation:

Given Integral 

Let, sin^-1 x = t ⇒  as we know the derivative of sin-1 x  =  

Now the equation reduces to 

 ⇒ we know ∫e^xdx = e^x + c 

∴ ∫ etdt = et + c, as t = sin-1 x our equation becomes e^sin-1x  + c

∴  = 

Given,

I = 

⇒ I = 

⇒ I =  = I₁ - I₂ (let)

Computing I₁,

I₁ = 

Put sin x = t → cos x dx = dt

⇒ I₁ = 

⇒ I1 = 

⇒ I1 = 

Similarly, 

I2 = 

Explanation:

 is not true for n = -1

because for n = -1

 =  = log |x| + c

(5) is correct

Explanation:

Let, t = x² ⇒ dt = 2xdx ⇒ xdx = dt/2

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

Substitude e^-x = t

-e^-x dx = dt