Fractional Calculations

Calculus Level 3

$\Large \int_{-1}^{2} \left |x^2 - \dfrac{1}{4} \right |dx = ?$

Notation: $|\cdot|$ denotes the absolute value function .

\dfrac{29}{12}

\dfrac{27}{12}

\dfrac{31}{12}

\dfrac{33}{12}

1 solution

Chew-Seong Cheong
Mar 3, 2018

$\begin{aligned} I & = \int_{-1}^2 \left| x^2 - \frac 14 \right| dx \\ & = {\color{#3D99F6}\int_{-1}^0 \left| x^2 - \frac 14 \right| dx} + \int_0^2 \left| x^2 - \frac 14 \right| dx & \small \color{#3D99F6} \text{Note that } \left| x^2 - \frac 14 \right| \text{ is even.} \\ & = {\color{#3D99F6}\int_0^1 \left| x^2 - \frac 14 \right| dx} + \int_0^2 \left| x^2 - \frac 14 \right| dx \\ & = 2\int_0^1 \left| x^2 - \frac 14 \right| dx + \int_1^2 \left| x^2 - \frac 14 \right| dx \\ & = 2\int_0^\frac 12 \left( - x^2 + \frac 14 \right) dx+ 2 \int_\frac 12^1 \left( x^2 - \frac 14 \right) dx + \int_1^2 \left( x^2 - \frac 14 \right) dx \\ & = 2\left[-\frac {x^3}3 +\frac x4\right]_0^\frac 12 + 2\left[\frac {x^3}3 -\frac x4\right]_\frac 12^1 + \left[\frac {x^3}3 -\frac x4\right]_1^2 \\ & = \frac 16 + \frac 13 + \frac {25}{12} = \boxed{\dfrac {31}{12}} \end{aligned}$

So the answers are adapted. It showed 31/24 first time. Yet somehow I clicked 27/24 lol. Anyhow I split the integral in - 1 to - 1/2 and 1/2 to 2. From - 1/2 to 1/2 is multiplied the given function by - 1 to cancel out the absolute value.

Peter van der Linden - 3 years, 3 months ago

Fractional Calculations

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