Question

$\displaystyle \int_{0}^{1} x\,dx + \int_{1}^{2} (2-x)\,dx = ...........$

Hint:

Piecewise functions can be integrated by evaluating each region separately.

Answer 1

Correct Answer: 1

Step 1: Evaluate first integral.
\[ \int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2} \] Step 2: Evaluate second integral.
\[ \int_1^2 (2-x)\,dx = \left[2x - \frac{x^2}{2}\right]_1^2 \] Substitute: At 2: $4 - 2 = 2$ At 1: $2 - \tfrac{1}{2} = \tfrac{3}{2}$ Difference: \[ 2 - \frac{3}{2} = \frac{1}{2} \] Step 3: Add both integrals.
\[ \frac{1}{2} + \frac{1}{2} = 1 \] Step 4: Conclusion.
Total value = 1.