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Question
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
Solution
\[f'\left( x \right) = x + b, f\left( 1 \right) = 5, f\left( 2 \right) = 13\]
\[ f'\left( x \right) = x + b\]
\[\int{f}'\left( x \right)dx = \int\left( x + b \right)dx\]
\[f\left( x \right) = \frac{x^2}{2} + bx + C . . . . (i)\]
\[f\left( 1 \right) = 5, f\left( 2 \right) = 13 \left( Given \right)\]
\[\text{Puting x} = \text{1 in (i)}\]
\[f\left( 1 \right) = \frac{1^2}{2} + b1 + C\]
\[5 = \frac{1}{2} + b + C . . . \left( ii \right)\]
\[\text{Puting x }= \text{2 in (i)}\]
\[f\left( 2 \right) = \frac{2^2}{2} + b2 + C\]
\[13 = \frac{4}{2} + 2b + C\]
\[13 = 2 + 2b + C . . . (iii)\]
\[\text{Solving (ii) and (iii) we get}, \]
\[b = \frac{13}{2} \text{and C }= - 2\]
\[Thus, f\left( x \right) = \frac{x^2}{2} + \frac{13}{2}x - 2\]
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