Evaluate the integral.

{eq}\displaystyle \int_0^2 \sqrt {|x - 1|}\ dx {/eq}


Evaluate the integral.

{eq}\displaystyle \int_0^2 \sqrt {|x - 1|}\ dx {/eq}

Integral of an Absolute Value Function:

Evaluating definite integrals of absolute value functions is a bit tricky as it does not follow the usual way of evaluating definite integrals by immediately finding antiderivatives and then utilizing the fundamental theorem of calculus. Instead, it first removes the absolute value symbol by affixing the appropriate sign to the function over the right portion of the integral interval.

Answer and Explanation: 1

The absolute value function {eq}|x-1| {/eq} is equal to {eq}x-1 {/eq} when {eq}x-1 \geq 0 \implies x \geq 1 {/eq}.

Otherwise, it is equal to {eq}-(x-1) = -x+1 {/eq}.

The portion of the interval where {eq}x \geq 1 {/eq} is {eq}[1,2] {/eq}, so the integral over that interval will be {eq}\displaystyle \int_1^2 \sqrt {x - 1} \ \mathrm{d}x {/eq}.

Meanwhile, the integral is {eq}\displaystyle \int_0^1 \sqrt {-x + 1} \ \mathrm{d}x {/eq} over the other portion of the interval, giving us:

{eq}\begin{align*} \\ \displaystyle \int_0^2 \sqrt {|x - 1|} \ \mathrm{d}x& =\int_0^1 \sqrt {-x + 1} \ \mathrm{d}x+\int_1^2 \sqrt {x - 1} \ \mathrm{d}x\\\\ & =\left[ - \frac{2(-x+1)^{\frac{3}{2}}}{3} \right]_{0}^{1}+\left[ \frac{2(x-1)^{\frac{3}{2}}}{3} \right]_{1}^{2} \\\\ & =\frac{2}{3}+\frac{2}{3} \\\\ & = \frac{4}{3}\\\\ \end{align*} {/eq}

Therefore, {eq}\displaystyle \int_0^2 \sqrt {|x - 1|} \ \mathrm{d}x= \frac{4}{3} {/eq}.

Learn more about this topic:

Absolute Value Function: Definition & Examples


Chapter 25 / Lesson 5

Absolute value definition. Learn the absolute value function equation, the domain of absolute value function, and see some absolute value function examples.

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