Find {eq}\displaystyle f {/eq} for the following function.

{eq}\displaystyle f'(x) = \frac{(x+1)}{\sqrt{x}}, \ f(1) = 5 {/eq}


Find {eq}\displaystyle f {/eq} for the following function.

{eq}\displaystyle f'(x) = \frac{(x+1)}{\sqrt{x}}, \ f(1) = 5 {/eq}

Initial Value Problem:

  • When we integrate the derivative of a function, we get back the antiderivative function. This is because integration and differentiation are reverse processes. In an initial value problem, the value of the function at a point is also given. We can use this initial value in the lower limits of integration. The upper limits of integration are to be kept in the variable form.
  • We shall be applying the following formulas:

{eq}\begin{align} \hspace{1cm} \int kdx&=kx+C & \left[ \text{Where C is an arbitrary constant of indefinite integration}~ \right] \\[0.3cm] \hspace{1cm} \int x^n\, dx&=\frac { x^{n+1}} {n+1}+C & \left[\text{ This is the power rule of integration } \right]\\[0.3cm] \end{align} {/eq}

Answer and Explanation: 1

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We have the following given data

{eq}\begin{align} f'(x) &= \frac{(x+1)}{\sqrt{x}} \\[0.3cm] f(1) & =5 \\[0.3cm] f (x)...

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Learn more about this topic:

Initial Value in Calculus: Definition, Method & Example


Chapter 11 / Lesson 13

Learn to define the initial value problem and initial value formula. Learn how to solve initial value problems in calculus. See examples of initial value problems.

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