Definite Integrals MCQ - 1

# Definite Integrals MCQ - 1

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Definite Integrals MCQ - 1 - Question 1

### Let f: be defined as f (t)  = f(t )dt

Definite Integrals MCQ - 1 - Question 2

### A minimum value of is

Detailed Solution for Definite Integrals MCQ - 1 - Question 2

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

Definite Integrals MCQ - 1 - Question 3

### Let f(x) = ∫ex(x−1)(x−2)dx then f decrease in the interval

Detailed Solution for Definite Integrals MCQ - 1 - Question 3

f(x) = ∫ex(x−1)(x−2)dx

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

⇒ex(x−1)(x−2) < 0

⇒(x−1)(x−2) < 0

⇒ ex(x−1)(x−2) < 0

⇒ (x−1)(x−2) < 0
⇒1 < x < 2,

∵ e> 0 ∀ x ∈ R

Definite Integrals MCQ - 1 - Question 4

Evaluate: Detailed Solution for Definite Integrals MCQ - 1 - Question 4 Definite Integrals MCQ - 1 - Question 5

Let f(x) = Then the real roots of tlie equation x2 — f'(x) = 0 are

Definite Integrals MCQ - 1 - Question 6 is equal to

Detailed Solution for Definite Integrals MCQ - 1 - Question 6

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)

Definite Integrals MCQ - 1 - Question 7

If f(x) is differentiable and equals

Definite Integrals MCQ - 1 - Question 8

If Definite Integrals MCQ - 1 - Question 9

The area of the region between the curves y = and bounded by the lines x = 0 and Definite Integrals MCQ - 1 - Question 10

Evaluate ∫73sin(t)-2cos(t)dt.

Detailed Solution for Definite Integrals MCQ - 1 - Question 10

Explanation ∫73sin(t)-2cos(t)dt = (cos(t)−2sin(t))73
= (cos(7) – 2sin(7)) – (cos(3) – 2sin(3))
= cos(7) – 2sin(7) – (cos(3) + 2sin(3)

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