4
Evaluating the Definite Integral of an Absolute Value Function
We are asked to evaluate the definite integral of the absolute value function \(f(x) = \left| {1 - {x^2}} \right|\) over the interval \([-2, 2]\). Evaluating a definite integral involving an absolute value function requires understanding where the expression inside the absolute value changes sign.
Understanding the Absolute Value Function
The function inside the absolute value is \(1 - x^2\). We need to determine where this expression is positive, negative, or zero within the interval \([-2, 2]\). The expression \(1 - x^2\) equals zero when \(x^2 = 1\), which means \(x = 1\) or \(x = -1\).
These points, \(x = -1\) and \(x = 1\), divide the interval \([-2, 2]\) into sub-intervals where the sign of \(1 - x^2\) is constant:
- For \(x \in (-1, 1)\), \(x^2 < 1\), so \(1 - x^2 > 0\). Thus, \(\left| {1 - {x^2}} \right| = 1 - x^2\) in this interval.
- For \(x \in (-2, -1)\), \(x^2 > 1\), so \(1 - x^2 < 0\). Thus, \(\left| {1 - {x^2}} \right| = -(1 - x^2) = x^2 - 1\) in this interval.
- For \(x \in (1, 2)\), \(x^2 > 1\), so \(1 - x^2 < 0\). Thus, \(\left| {1 - {x^2}} \right| = -(1 - x^2) = x^2 - 1\) in this interval.
At the points \(x = -1\) and \(x = 1\), \(1 - x^2 = 0\), and \(\left| {1 - {x^2}} \right| = 0\).
Splitting the Definite Integral
Because the definition of the absolute value function changes at \(x = -1\) and \(x = 1\), we must split the original definite integral into a sum of definite integrals over the sub-intervals:
\(\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^{ - 1} {\left| {1 - {x^2}} \right|} dx + \int\limits_{ - 1}^1 {\left| {1 - {x^2}} \right|} dx + \int\limits_1^2 {\left| {1 - {x^2}} \right|} dx\)
Now, substitute the appropriate expression for \(\left| {1 - {x^2}} \right|\) in each integral based on the sign analysis:
\(\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^{ - 1} {(x^2 - 1)} dx + \int\limits_{ - 1}^1 {(1 - x^2)} dx + \int\limits_1^2 {(x^2 - 1)} dx\)
Evaluating Each Segment of the Integration
We now evaluate each definite integral separately using the fundamental theorem of calculus. The antiderivative of \(x^2 - 1\) is \(\frac{x^3}{3} - x\), and the antiderivative of \(1 - x^2\) is \(x - \frac{x^3}{3}\).
First Integral:
\(\int\limits_{ - 2}^{ - 1} {(x^2 - 1)} dx = \left[ \frac{x^3}{3} - x \right]_{-2}^{-1}\)
\(= \left(\frac{(-1)^3}{3} - (-1)\right) - \left(\frac{(-2)^3}{3} - (-2)\right)\)
\(= \left(-\frac{1}{3} + 1\right) - \left(-\frac{8}{3} + 2\right)\)
\(= \left(\frac{-1 + 3}{3}\right) - \left(\frac{-8 + 6}{3}\right)\)
\(= \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}\)
Second Integral:
\(\int\limits_{ - 1}^1 {(1 - x^2)} dx = \left[ x - \frac{x^3}{3} \right]_{-1}^1\)
\(= \left(1 - \frac{1^3}{3}\right) - \left((-1) - \frac{(-1)^3}{3}\right)\)
\(= \left(1 - \frac{1}{3}\right) - \left(-1 - \left(-\frac{1}{3}\right)\right)\)
\(= \frac{2}{3} - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}\)
Third Integral:
\(\int\limits_1^2 {(x^2 - 1)} dx = \left[ \frac{x^3}{3} - x \right]_1^2\)
\(= \left(\frac{2^3}{3} - 2\right) - \left(\frac{1^3}{3} - 1\right)\)
\(= \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right)\)
\(= \left(\frac{8 - 6}{3}\right) - \left(\frac{1 - 3}{3}\right)\)
\(= \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}\)
The total definite integral is the sum of these results:
\(\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|} dx = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4\)
This calculation shows how to handle the absolute value function in a definite integral by splitting the integration interval based on the sign changes of the expression inside the absolute value. This application of calculus is crucial for evaluating integrals of piecewise functions.
| Interval | Function | Definite Integral | Value |
|---|---|---|---|
| \([-2, -1]\) | \(x^2 - 1\) | \(\int\limits_{ - 2}^{ - 1} {(x^2 - 1)} dx\) | \(\frac{4}{3}\) |
| \([-1, 1]\) | \(1 - x^2\) | \(\int\limits_{ - 1}^1 {(1 - x^2)} dx\) | \(\frac{4}{3}\) |
| \([1, 2]\) | \(x^2 - 1\) | \(\int\limits_1^2 {(x^2 - 1)} dx\) | \(\frac{4}{3}\) |
The total definite integral value is the sum of the values from each segment.
Total definite integral \( = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4 \).
Understanding this process is key to mastering definite integral problems involving absolute values in calculus.
Important Questions from Indefinite Integrals
The value of \(\rm \int \frac{(x+1)}{x(xe^x + 1)}\ dx\) is equal to:
What is ∫ (e log x + sin x) cos x dx equal to?
Evaluation of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\) is:
The value of \(\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \) is:
The value of \(\int_0^\pi {{{\sin }^6}} x \cdot {\cos ^5}x dx\) is:
