Again Calc ??

Calculus Level 3

Let $$f(x) = x^2$$ ; $$0\leq x < 1$$ and $$f(x) = \sqrt{x}$$ ; $$1\leq x \leq 2$$

find $$\displaystyle\int_0^2 \ f(x)\ dx$$

answer is $$\dfrac{a\sqrt{b}-1}{3}$$ find $$a+b$$

$$a$$ and $$b$$ are positive integers, and $$b$$ is not divisible by the square of any prime.

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