Let [[x]] denote the greatest integer [ x].

Question

Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx} $$ is

Accepted Answer

Combining both the graph we get,

Both graph cuts each other at $y=2$ and $y=1$

$$\therefore x^2=2$$

$$\Rightarrow x=\sqrt2$$

Two graphs cuts each other at $x=1$ and $$x=\sqrt2$$

In 0 to 1, $$f(x) = \max ({x^2},1 + [x]) = 1$$

In 1 to $$\sqrt2$$, $$f(x) = \max ({x^2},1 + [x]) = 2$$

In $$\sqrt2$$ to 2, $$f(x) = \max ({x^2},1 + [x]) = {x^2}$$

$$\therefore$$ $$\int\limits_0^2 {f(x)dx} $$

$$ = \int\limits_0^1 {1\,dx + \int\limits_1^{\sqrt 2 } {2\,dx + \int\limits_{\sqrt 2 }^2 {{x^2}\,dx} } } $$

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

$$ = 1 + 2(\sqrt 2  - 1) + {8 \over 3} - {{2\sqrt 2 } \over 3}$$

$$ = {8 \over 3} - 1 + 2\sqrt 2  - {{2\sqrt 2 } \over 3}$$

$$ = {5 \over 3} + {{6\sqrt 2  - 2\sqrt 2 } \over 3}$$

$$ = {5 \over 3} + {{4\sqrt 2 } \over 3}$$

$$ = {{5 + 4\sqrt 2 } \over 3}$$

Solutions · 1