Question

For the following exercises, solve the initial value problem. 499. $f'(x) = x^{-3}$, $f(1) = 1$ 500. $f'(x) = \sqrt{x} + x^2$, $f(0) = 2$ 501. $f'(x) = \cos x + \sec^2(x)$, $f(\frac{\pi}{4}) = 2 + \frac{\sqrt{2}}{2}$ 502. $f'(x) = x^3 - 8x^2 + 16x + 1$, $f(0) = 0$

          For the following exercises, solve the initial value problem.
499. $f'(x) = x^{-3}$, $f(1) = 1$
500. $f'(x) = \sqrt{x} + x^2$, $f(0) = 2$
501. $f'(x) = \cos x + \sec^2(x)$, $f(\frac{\pi}{4}) = 2 + \frac{\sqrt{2}}{2}$
502. $f'(x) = x^3 - 8x^2 + 16x + 1$, $f(0) = 0$

For the following exercises, solve the initial value problem.
499. f'(x) = x^-3, f(1) = 1
500. f'(x) = √(x) + x^2, f(0) = 2
501. f'(x) = cos x + sec^2(x), f((π)/(4)) = 2 + (√(2))/(2)
502. f'(x) = x^3 - 8x^2 + 16x + 1, f(0) = 0

Added by Rebecca H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

11/02/2023

Step 1

Step 1: Find the antiderivative of $f'(x)$. Show more…

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For the following exercises, solve the initial value problem. 499. $f'(x) = x^{-3}$, $f(1) = 1$ 500. $f'(x) = \sqrt{x} + x^2$, $f(0) = 2$ 501. $f'(x) = cosx + sec^2(x)$, $f(\frac{\pi}{4}) = 2 + \frac{\sqrt{2}}{2}$ 502. $f'(x) = x^3 - 8x^2 + 16x + 1$, $f(0) = 0$