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  • If the function <img alt="f:[0,8] \rightarrow R" src="https://entrancecorner.

If the function f:[0,8] \rightarrow R  is differentiable then for 0<\alpha, \beta<2, \int_0^8 f(t) d t

is equal to

Option: 1

3\left[\alpha^3 f\left(\alpha^2\right)+\beta^3 f\left(\beta^2\right)\right]


Option: 2

3\left[\alpha^3 f(\alpha)+\beta^3 f(\beta)\right]


Option: 3

3\left[\alpha^2 f\left(\alpha^3\right)+\beta^2 f\left(\beta^3\right)\right]


Option: 4

3\left[\alpha^2 f\left(\alpha^2\right)+\beta^2 f\left(\beta^2\right)\right]


Answers (1)

best_answer

Let g(x)=\int_0^{x^3} f(t) d t

Now \int_0^8 f(t) d t=g(2)=\frac{g(2)-g(1)}{2-1}+\frac{g(1)-g(0)}{1-0}=g^{\prime}(\alpha)+g^{\prime}(\beta)=3\left[\alpha^2 f\left(\alpha^3\right)+\beta^2 f\left(\beta^3\right)\right].

Posted by

seema garhwal

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