Let f: R R be a twice differentiable function satisfying f(2) = - 1, f '(2) = 4 and the value of f(3) is equal to

JEE MainMathematicsEasyDefinite Integration

Let f: R $\rightarrow$ R be a twice differentiable function
satisfying f(2) = - 1, f '(2) = 4 and
$\int_{2}^{3}\mspace{2mu}(3 - x)f^{''}(x)dx = 7$ the value of f(3) is
equal to

Show answer and solution

\because\int_{2}^{3}\mspace{2mu}(3 - x)f^{''}(x)dx = 7

\Rightarrow \left\lbrack (3 - x)f^{'}(x) \right\rbrack_{2}^{3} - \int_{2}^{3}\mspace{2mu}( - 1)f^{'}(x)dx = 7

0 - f^{'}(2) + \int_{2}^{3}\mspace{2mu} f^{'}(x)dx = 7

f(3) = 10

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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