Maximising integrals

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 $$a,b$$.

Note by Paramjit Singh
4 years, 4 months ago

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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

- 4 years, 1 month ago

- 4 years, 1 month ago

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.

- 4 years, 4 months ago

Hint : The integrand is positive only between two values that are its roots, take them as a and b to maximise the integral.

- 4 years, 4 months ago

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

- 4 years, 4 months ago