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The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x =1$ form an angle of $\pi/6$ and at the point $x =2$ , an angle of $\pi/3$ and at the point $x =3$ , an angle of $\pi/4$ .The value of $\int_{1}^{3}{f}'(x){f}''(x)dx+\int_{2}^{3}{f}''(x)dx$ . $({f}''(x)$ is supposed to be continuous) is :
Option: 1
$\frac{4\sqrt{3}-1}{3\sqrt{3}}$

Option: 2
$\frac{3\sqrt{3}-1}{2}$

Option: 3
$\frac{4-\sqrt{3}}{3}$

Option: 4
none of these

Answers (1)

best_answer

lower and upper limit -

$\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}$

$= F\left ( b \right )-F\left ( a \right )$

- wherein

Where a is lower and b is upper limit.

Integration By PARTS -

Let $u$ and $v$ are two functions then

$\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx$

- wherein

Where $u$ is the Ist function $v$ is he IInd function

Given ,

$f'(1)=\left ( \frac{dy}{dx} \right )_{x=1}=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$

$f'(2)=\left ( \frac{dy}{dx} \right )_{x=2}=\tan\frac{\pi}{3}=\sqrt{3}$

$f'(3)=\left ( \frac{dy}{dx} \right )_{x=3}=\tan\frac{\pi}{4}=1$

Let , $I=\int_{1}^{3}f'(x).f''(x)dx+\int_{2}^{3}f''(x)dx=I_{1}+I_{2}$

$\therefore I_{1}=\int_{1}^{3}f'(x).f''(x)dx$

$I_{1}=f'(x).f'(x) | \right |_{1}^{3}- \int_{1}^{3}f''(x)f'(x)dx$

$2I_{1}=\left \{ f'(3) \right \}^{2}-\left \{ f'(1) \right \}^{2}$

$2I_{1}=1-\frac{1}{3}$

$I_{1}=\frac{1}{3}$

$I_{2}=\int_{2}^{3}f''(x)dx=f'(x) | \right |^{3}_{2}=f'(3)-f'(2)$

$=1-\sqrt{3}$

$\therefore I=I_{1}+I_{2}=\frac{1}{3}+1-\sqrt{3}=\frac{4}{3}-\sqrt{3}$

Posted by

Deependra Verma

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