Definite Integrals MCQ - 1
10 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Definite Integrals MCQ - 1
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be defined as f (t) = 
f(t )dt
isCorrect 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
Let f(x) = ∫ex(x−1)(x−2)dx then f decrease in the interval
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,
∵ ex > 0 ∀ x ∈ R
Let f(x) = 
Then the real roots of tlie equation x2 — f'(x) = 0 are
is equal toANSWER :- 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)
equals
The area of the region between the curves y = 
and bounded by the lines x = 0 and 
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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