Match List I to List II. (P) ∫ 0 1 [ 1 − x 2 ( 1 + 1 − x 2 ) ] − 1 d x . (Q) If f ( x ) = x 5 + e x /2 + e x 3 and g = f − 1 , find g ′ ( 2 ) . (R) lim n → ∞ n 1 ∑ r = 1 4 n r 1 . (S

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Match List I to List II. (P) ∫ 0 1 ​ [ 1 − x 2 ​ ( 1 + 1 − x 2 ​ ) ] − 1 d x . (Q) If f ( x ) = x 5 + e x /2 + e x 3 and g = f − 1 , find g ′ ( 2 ) . (R) lim n → ∞ ​ n ​ 1 ​ ∑ r = 1 4 n ​ r ​ 1 ​ . (S) If f ( x ) = ∫ x − 1 2 x ​ ( t 3 − 2 t + 4 ) d t , find f ′ ( 1 ) . List II: (1) 2, (2) 4, (3) 12, (4) 1.

TestHub · Accepted Answer

(P) Put x = sin θ to obtain ∫ 0 π /2 ​ d θ / ( 1 + cos θ ) = 1 , P-4. (Q) f ( 0 ) = 2 and f ′ ( 0 ) = 1/2 , so g ′ ( 2 ) = 2 , Q-1. (R) Rewrite it as n − 1 ∑ r = 1 4 n ​ ( r / n ) − 1/2 ; the limit is ∫ 0 4 ​ t − 1/2 d t = 4 , R-2. (S) Leibniz gives f ′ ( 1 ) = 12 , S-3. The source key is (A), but item R needs visual confirmation.

Mathematics - Definite Integration Question with Solution | TestHub

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