Let y=∫ (u^3)/(1+u2)du from 1-4x to 1. Use the Fundamental Theorem of Calculus to find y′?
Let y=∫ (u^3)/(1+u2)du from 1-4x to 1. Use the Fundamental Theorem of Calculus to find y′
I get -4[ ((1^3)/(1+1) - (1-4x)^3/(1+(1-4x)^2) ]
but the correct answer is -4(-(1-4x)^3/(1+(1-4x)^2)) which completely ignores the upper limit 1 and only looks at lower limit 1-4x.
Can any1 explain how to do this question??
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1) ∫ (u^3)/(1+u2)du from 1-4x to 1 = (1/2)*{1 - ln(2)} - (1/2)*{(1-4x)² - lnI1 + (1-4x)²I}
2) Differentiating the above, y' = 0 - (1/2)*[2*(1-4x)*(-4) - (2)*(1-4x)(-4)/{1 + (1-4x)²}]
[Since first part {1 - ln(2)} is being constant, the differential of which = 0]
On simplifying this would become = (-4)*[-(1-4x)³/{1 + (1-4x)²}]; so the answer given is correct. You may verify your work out.
EDIT;
Add the following to the above solution in the beginning.
i) I suppose you have directly evaluated this (u^3)/(1+u2) for upper limit 1 and lower limit (1-4x) and obtained your answer as -4[ ((1^3)/(1+1) - (1-4x)^3/(1+(1-4x)^2) ], which is not correct.
ii) You are supposed to initially integrate the function, evaluate for upper and lower limits, get the function and then find its differential.
iii) u³/(1+u²) = u - {u/(1+u²)}, the integral of which = (u²/2) - (1/2)*lnI1+u²I
Evaluating this for its upper limit 1 and lower limit (1-4x), you get the following: