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.

1 solution

Tín Phạm Nguyễn
Mar 13, 2015

As $\displaystyle\lim_{x \to 1^-} f(x) = \displaystyle\lim_{x \to 1^+} f(x) = 1$ , $f(x)$ is continuous in $[0,2]$ .

Hence:

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

$= \displaystyle\int_0^1 \ f(x) \ dx + \displaystyle\int_1^2 \ f(x) \ dx$

$= \displaystyle\int_0^1 \ x^2 \ dx + \displaystyle\int_1^2 \sqrt {x} \ dx\$

$= \left[ \displaystyle\frac {x^3} {3} \right]_0^1 + \left[ \displaystyle\frac {2x \sqrt {x}} {3} \right]_1^2$

$= \displaystyle\frac {1} {3} - 0 + \displaystyle\frac {4 \sqrt {2}} {3} - \frac {1} {3}$

$= \displaystyle\frac {4 \sqrt {2} - 1} {3}$

Hence, $a + b = 4 + 2 = 6$