Correct Answer - Option 3 : 2
Concept:
Definite Integral:
If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = \left[ g(x)\right]_a^b\) = g(b) - g(a).
Properties:
- For even functions: f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
- For odd functions: f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
-
\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
- If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).
Calculation:
It can be observed that sin (-θ) = -sin θ.
Let f(x) = |sin x|
Put x = -x
⇒ f(-x) = |sin -x| = |-sin x| = |sin x| = f(x)
∴ |sin x| is an even function.
Therefore, using the properties of definite integrals, we get:
\(\rm \int_{-\pi/2}^{\pi/2}|\sin x|\ dx\)
= 2\(\rm \int_{0}^{\pi/2}\sin x\ dx\)
= 2\(\rm \left[-\cos x\right]_{0}^{\pi/2}\)
= -2\(\rm \left[\cos \pi/2-\cos 0\right]\)
= -2[0 - 1]
= 2.
