The equation of the curve y = f ( x ) .The tangents at :-
(1, f ( 1 ) ),(2, f ( 2 ) ) ,and(3, f ( 3 ) ) makes angle 6 π , 3 π and 4 π respectively with the positive direction of x - axis.Then the value of
∫ 2 3 f ′ ( x ) f " ( x ) d x + ∫ 1 3 f " ( x ) d x can be expressed as
a − 1 find the value of a .
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Nice solution sir.
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tailored the original question?? hun?
Thanks!! Just wanna increase the number of followers!!
Well not a level 5 I suppose!!!!
If we look at the first integrand, we see an f ′ ( x ) and an f ′ ′ ( x ) . This should remind us of chain rule, and we should think about how we would differentiate a function to get f ′ ( x ) and f ′ ′ ( x ) . We should realize that we need to differentiate f ′ ( x ) to some power in order to obtain the f ′ ′ ( x ) , and then we realize that power has to be 2.
The indefinite integral of ∫ f ′ ( x ) f ′ ′ ( x ) = 2 1 f ′ ( x ) 2
Just by simple integration, we see that the indefinite integral of the other integrand is f ′ ( x ) .
We have bounds given to us, so we now need to find the definite values of these indefinite integrals. If the tangent at (1, f(1)) makes an angle of pi/6 with the +x axis, we know the slope, and therefore the quantity f ′ ( 1 ) , is arctan(pi/6)= 3 1 . The same principle can be applied to the other two points, and we find f ′ ( 2 ) = arctan 3 π = 3 and f ′ ( 3 ) = arctan 4 π = 1 .
Then we go to our integrals and plug in our values:
∫ 2 3 f ′ ( x ) f ′ ′ ( x ) = 2 1 ( f ′ ( 3 ) ) 2 − 2 1 ( f ′ ( 2 ) ) 2
and
∫ 1 3 f ′ ′ ( x ) = f ′ ( 3 ) − f ′ ( 1 )
Plugging in our values and adding the two integrals we get it to come out to 3 − 1 , so our answer is a = 3 ≈ 1 . 7 3 2 .
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It is given that:
f ′ ( 1 ) = tan 6 π = 3 1 ⇒ f ′ ( 2 ) = tan 3 π = 3 ⇒ f ′ ( 3 ) = tan 4 π = 1
and we need to solve the integral below:
∫ 2 3 f ′ ( x ) f ′ ′ ( x ) d x + ∫ 1 3 f ′ ′ ( x ) d x
Now consider:
d x d ( f ′ ( x ) f ′ ( x ) ) = f ′ ( x ) f ′ ′ ( x ) + f ′ ′ ( x ) f ′ ( x ) = 2 f ′ ( x ) f ′ ′ ( x )
⇒ ∫ f ′ ( x ) f ′ ′ ( x ) d x = 2 1 f ′ ( x ) f ′ ( x ) + C
It is obvious that: ∫ f ′ ′ ( x ) d x = f ′ ( x ) + C
Therefore,
∫ 2 3 f ′ ( x ) f ′ ′ ( x ) d x + ∫ 1 3 f ′ ′ ( x ) d x = [ 2 1 f ′ ( x ) f ′ ( x ) ] x = 2 x = 3 + [ f ′ ( x ) ] x = 1 x = 3
= 2 1 ( 1 − 3 ) + 1 − 3 1 = 3 − 1 ⇒ a = 3