Definite Integrals MCQ - 1 - Mathematics MCQ

Correct Answer :- d

Explanation : f(x) = ∫(0 to x) te^{-t^2} dt

f'(x) = xe^{-x^2} = 0

Therefore, x = 0

f"(x) = e^{-x^2}(1 - 2x²)

f"(0) = 1 > 0

Therefore, the minimum value is f(0) = 0

f(x) = ∫e^x(x−1)(x−2)dx

For decreasing function ,f′(x)<0,f′(x)<0

⇒e^x(x−1)(x−2) < 0

⇒(x−1)(x−2) < 0

⇒ e^x(x−1)(x−2) < 0

⇒ (x−1)(x−2) < 0
⇒1 < x < 2,

∵ e^x > 0 ∀ x ∈ R

f(x) = ∫ (from 1 to x) √(2 − t²) dt  
f'(x) = √(2 − x²)  
x² − f'(x) = 0  
x² = √(2 − x²)  
⇒ x⁴ + x² − 2 = 0  
x² = 1, also  
x² = −2  
the real roots are  
x² = 1  
⇒ x = ±1  
 

ANSWER :- A

Solution :- limx→0 1/x[∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]

 limx→0  [∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]/x

Hence it is 0/0 form, Apply L-hospital rule.

lim x-->0  0 - {e^(sin^2a).0 - e^(sin^2(x+y).1}/1

lim x-->0 e^(sin^2(x+y))

= e^(sin^2y)

• Understand the Given Equation:

  We are given the equation:

  Integral from sin(x) to 1 of [t² * f(t)] dt = 1 - sin(x)

  This equation holds true for every x between 0 and π/2.

• Differentiate Both Sides with Respect to x:

  To find the function f(t), we'll differentiate both sides of the equation with respect to x.

  Let F(x) represent the left side of the equation:

  F(x) = Integral from sin(x) to 1 of [t² * f(t)] dt

  Differentiating F(x) with respect to x using Leibniz's Rule (which allows differentiation under the integral sign), we get:

  F'(x) = -cos(x) * [sin(x)]² * f(sin(x))

  Here, the negative sign comes from differentiating the lower limit sin(x), and cos(x) is the derivative of sin(x).

• Differentiate the Right Side:

  The right side of the original equation is:

  1 - sin(x)

  Differentiating this with respect to x gives:

  d/dx (1 - sin(x)) = -cos(x)

• Set the Derivatives Equal:

  Since both sides of the original equation are equal for all x in the given interval, their derivatives must also be equal. Therefore:

  -cos(x) * [sin(x)]² * f(sin(x)) = -cos(x)

  We can simplify this by canceling out the -cos(x) term from both sides (assuming cos(x) is not zero, which it isn't in the interval from 0 to π/2):

  [sin(x)]² * f(sin(x)) = 1

• Solve for f(sin(x)):

  From the simplified equation:

  [sin(x)]² * f(sin(x)) = 1

  We can solve for f(sin(x)):

  f(sin(x)) = 1 / [sin(x)]²

• Express f(t) in Terms of t:

  Let t be equal to sin(x). Since x is in the interval from 0 to π/2, t ranges from 0 to 1.

  Therefore, the function f(t) can be expressed as:

  f(t) = 1 / t²

1. Compute f at 1 divided by the Square Root of 3:

   Finally, we substitute t with 1 divided by the square root of 3:

   f(1/√3) = 1 / (1/√3)²

   Simplifying the denominator:

   (1/√3)² = 1/3

   Therefore:

   f(1/√3) = 1 / (1/3) = 3

We know that cot² x = cosec²x – 1

∫cot²x dx = ∫ (cosec²x – 1) dx = -cot x -x + C. [Since, ∫cosec²x dx = -cot x + c]

Hence, the correct answer is option (b) -cot x – x + C.