First slide

Evaluation of definite integrals

Question

A line tangent to the graph of the function $y = f (x)$ at the point $x = a$ forms an $π / 3$ with $y -$ axis and at $x = b$ an angle $π / 4$ with $x -$ axis then
$\int_{a}^{b} f^{''} (x) d x$ is

Moderate

A

$1 - \frac{1}{\sqrt{3}}$
B

$- \frac{π}{12}$
C

$\frac{π}{12}$
D

$\sqrt{3} - 1$

Solution

$f^{'} (a) = \tan (\frac{π}{2} - \frac{π}{3}) = \tan \frac{π}{6} = \frac{1}{\sqrt{3}}$ and $f^{'} (b) =$
$\tan \frac{π}{4} = 1$ , so
$\int_{a}^{b} f^{''} (x) d x = f^{'} (b) - f^{'} (a) = 1 - \frac{1}{\sqrt{3}} .$

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