Suppose \(a<b\). Then find maximum value of the integral

\(\displaystyle \int_a^b (\frac{3}{4} - x - x^2) dx\) over all possible values of \(a,b\).

Suppose \(a<b\). Then find maximum value of the integral

\(\displaystyle \int_a^b (\frac{3}{4} - x - x^2) dx\) over all possible values of \(a,b\).

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{a}^{b}((3/4)-x-x²)dx=(d/(da)) ((1/3)a³+(1/2)a²-(3/4)a-(1/3)b³-(1/2)b²+(3/4)b)= a²+a-(3/4)=0, Solution is: (1/2),-(3/2) (d/(db)) ((1/3)a³+(1/2)a²-(3/4)a-(1/3)b³-(1/2)b²+(3/4)b)= -b²-b+(3/4)=0, Solution is: (1/2),-(3/2) ∫{-(3/2)}^{(1/2)}((3/4)-x-x²)dx= 1. 3333 – Carlos Suarez · 2 years, 8 months agoLog in to reply

Answer=1.3333333 – Carlos Suarez · 2 years, 8 months ago

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Sketch a graph and it becomes clear what are the values of a and b. Note that a definite integral is the area under the curve. – Pranav Arora · 2 years, 11 months ago

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Hint :The integrand is positive only between two values that are its roots, take them as a and b to maximise the integral. – Jatin Yadav · 2 years, 11 months agoLog in to reply

– Paramjit Singh · 2 years, 11 months ago

Find the area, my friend, I know how to do this. Find the answer. Never mind, thanks.Log in to reply