1.

Determine f(x) for f'(x) = \(\rm 4x^3 - 2 {1\over \sqrt x} + e^{-4x}\), and f(0) = \(3\over4\)1. x4 - 4\(\rm\sqrt x - {e^{-4x}\over4}\) + 12. x4 - 4\(\rm\sqrt x + {e^{-4x}\over4}\) + 13. x4 - \(\rm\sqrt x - {e^{-4x}\over4}\) + 14. x4 - 4\(\rm\sqrt x +e^{-4x}\) + 1

Answer» Correct Answer - Option 1 : x4 - 4\(\rm\sqrt x - {e^{-4x}\over4}\) + 1

Concept:

Integral property:

 
  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C
  • ∫ eax dx = \(\rm e^{ax}\over a\)+ C
  • ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C
 

Calculation:

Given f'(x) = \(\rm 4x^3 - 2 {1\over \sqrt x} + e^{-4x}\)

f(x) = ∫ f'(x) dx

⇒ f(x) = \(\rm \int 4x^3 - 2 {1\over \sqrt x} + e^{-4x}\) dx

⇒ f(x) = \(\rm 4\left[{x^4\over4}\right] - 2\left[{x^{1\over2}\over{1\over2}}\right] + \left[{e^{-4x}\over-4}\right] \)

⇒ f(x) = x4 - 4\(\rm\sqrt x - {e^{-4x}\over4}\) + C

Now f(0) = \(3\over4\)

⇒  04 - 4\(\rm\sqrt 0 - {e^{-4(0)}\over4}\) + C = \(3\over4\)

⇒ C - \(1\over4\) = \(3\over4\)

⇒ C = 1

⇒ f(x) = x4 - 4\(\boldsymbol{\rm\sqrt x - {e^{-4x}\over4}}\) + 1


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