Question

Find a function $f$that has the given derivative $f\text{'}$and value $f\left(c\right)$. Find an antiderivative of $f\text{'}$by hand, if possible; if it is not possible to antidifferentiation by hand, use the Second Fundamental Theorem of Calculus to write down an antiderivative.

$f^{\mathrm{\prime }}\left(x\right)=2\sin \left(\pi x\right),f\left(2\right)=4$

Short answer

Ans: The function is,$f\left(x\right)=\frac{1}{2}(\ln |2x-1\left|\right)+3$

Step-by-step solution

Step 1. Given information.

given,

$f^{\mathrm{\prime }}\left(x\right)=2\sin \left(\pi x\right),f\left(2\right)=4$

Step 2. The objective is to find a function f meeting the above values.

So,

$\begin{array}{r}f\left(x\right)=\int 2\sin \left(\pi x\right)dx\\ =\int \frac{1}{y}\frac{dy}{2}\\ =\frac{1}{2}(\ln |y\left|\right)+c\\ =\frac{1}{2}(\ln |2x-1\left|\right)+c\end{array}$

The function is, $\frac{1}{2}(\ln |2x-1\left|\right)+c$

Step 3. Finding the value of c,

$\begin{array}{r}f\left(1\right)=3\\ \frac{1}{2}(\ln |2\left(1\right)-1\left|\right)+c=3\\ \frac{1}{2}\ln 1+c=3\\ c=3\end{array}$

Therefore, the function is $f\left(x\right)=\frac{1}{2}(\ln |2x-1\left|\right)+3$.