QUIZ #11: CALCULUS 1A (Stankova) Wednesday, April 14, 2004 Section 10:00–11:00 (Voight)

Problem 1. Find the antiderivative F of f that satisfies the given condition.

f(x) = 4 − sec^{2}(2x), F (π) = 0.

Solution. We compute that

F (x) = 4x − tan(2x)/2 + C.

Since F (π) = 0, we have

F (π) = 4π − tan(2π)/2 + C = 4π − 0 + C = 0

so C = −4π. Therefore

F (x) = 4x − tan(2x)/2 − 4π.

QUIZ #11: CALCULUS 1A (Stankova) Wednesday, April 14, 2004 Section 11:00–12:00 (Voight)

Problem 1. Find f, if

f′′(x) = sin x + x^{−1/2}.

Solution. We have

f′(x) = − cos x + 2x^{1/2} + C

and so

f(x) = − sin x + 4x^{3/2}/3 + Cx + D.