Question

Let

Evaluate \(\int\limits_{{\rm{ - 3}}}^{\rm{1}} {{\rm{f(x)dx}}} \) by interpreting the integral as a difference of areas.

Short answer

Integral \(\int\limits_{ - 3}^1 {f(x)dx} \)as a difference of areas is \(\frac{{6 - \pi }}{4}\).

Step-by-step solution

Plotting the given function

The area above X-axis is positive and area below Y-axis is negative.

Considering the given values

For \( - 3 \le x \le 0\), \(f(x) = - x - 1\) and

For \(0 \le x \le 1\), \(f(x) = - \sqrt {1 - {x^2}} \)

Finding Area for given intervals

Let \({A_1}{\rm{ }}and{\rm{ }}{A_2}\)be the area of the triangles formed by a line.

\(\begin{aligned}{c}{A_1} &= \frac{1}{2}(2)(2)\\ &= 2\end{aligned}\)

\(\begin{aligned}{c}{A_2} &= \frac{1}{2}(1)( - 1)\\ &= \frac{{ - 1}}{2}\end{aligned}\)

Let \({A_3}\)be the area of the curve.

\(\begin{aligned}{c}y &= - \sqrt {1 - {x^2}} \\{y^2} &= 1 - {x^2}\end{aligned}\)

\(\therefore {x^2} + {y^2} = 1\)

This is the equation of circle centered at origin with radius 1.

\(\begin{aligned}{c}{A_3} &= \frac{{ - \pi {r^2}}}{4}\\ &= \frac{{ - 1}}{4} \cdot \pi \cdot {(1)^2}\\ &= - \frac{\pi }{4}\end{aligned}\)

Adding all together,

\(\begin{aligned}{c}A &= {A_1} + {A_2} + {A_3}\\ &= 2 - \frac{1}{2} - \frac{\pi }{4}\\ &= \frac{{8 - 2 - \pi }}{4}\\ &= \frac{{6 - \pi }}{4}\end{aligned}\)

Therefore, Integral \(\int\limits_{ - 3}^1 {f(x)dx} \)as a difference of areas is \(\frac{{6 - \pi }}{4}\).