The tangent to the graph of the function y - fx at the point with abscissa x - a forms with the x-axis an angle of -4 and at the point with abscissa x - b at an angle of -3- then find the value of b a - f- x - f- x dx

f ' (a) = 1 and f' #160; (b) =√(3) #160; ;Now I = [ f'(x)/2 ]_a^b=(f'(b))^2 (f'(a))^2/2 =3 1/2=1 ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

The tangent t...

Question

The tangent to the graph of the function y = f(x) at the point with abscissa x = a forms with the x-axis an angle of $\frac{π}{4}$ and at the point with abscissa x = b at an angle of $\frac{π}{3}$ , then find the value of
$b \int a f^{'} (x) . f^{''} (x) d x$

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Solution

$f^{'} (a) = 1$ and $f^{'} (b) = \sqrt{3}$ ;
Now $I = {[\frac{f^{'} (x)}{2}]}_{a}^{b} = \frac{(f^{'} (b))^{2} - (f^{'} (a))^{2}}{2}$
$= \frac{3 - 1}{2} = 1$

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The tangent to the graph of the function y = f(x) at the point with abscissa x = a forms with the x-axis an angle of π4 and at the point with abscissa x = b at an angle of π3, then find the value ofb∫af′(x).f′′(x)dx

f′(a)=1 and f′(b)=√3 ;Now I=[f′(x)2]ba=(f′(b))2−(f′(a))22=3−12=1

The tangent to the graph of the function y = f(x) at the point with abscissa x = a forms with the x-axis an angle of $\frac{π}{4}$ and at the point with abscissa x = b at an angle of $\frac{π}{3}$ , then find the value of
$b \int a f^{'} (x) . f^{''} (x) d x$

$f^{'} (a) = 1$ and $f^{'} (b) = \sqrt{3}$ ;
Now $I = {[\frac{f^{'} (x)}{2}]}_{a}^{b} = \frac{(f^{'} (b))^{2} - (f^{'} (a))^{2}}{2}$
$= \frac{3 - 1}{2} = 1$