this question is designed to be answered without a calculator. the area enclosed by y = x² - x and y = 1…

Explanation:

Step1: Find intersection points

Set $x^{2}-x = 1 - x^{2}$. [ \begin{align*} x^{2}-x-(1 - x^{2})&=0\ 2x^{2}-x - 1&=0\ (2x + 1)(x - 1)&=0 \end{align*} ] So $x=-\frac{1}{2}$ and $x = 1$.

Step2: Determine upper - lower functions

For $-\frac{1}{2}\leq x\leq1$, $1 - x^{2}\geq x^{2}-x$.

Step3: Calculate the area

The area $A=\int_{-\frac{1}{2}}^{1}[(1 - x^{2})-(x^{2}-x)]dx=\int_{-\frac{1}{2}}^{1}(1 - 2x^{2}+x)dx$. [ \begin{align*} \int_{-\frac{1}{2}}^{1}(1 - 2x^{2}+x)dx&=\left[x-\frac{2}{3}x^{3}+\frac{1}{2}x^{2}\right]_{-\frac{1}{2}}^{1}\ &=(1-\frac{2}{3}+\frac{1}{2})-(-\frac{1}{2}+\frac{2}{3}\times\frac{1}{8}+\frac{1}{2}\times\frac{1}{4})\ &=(1-\frac{2}{3}+\frac{1}{2})-(-\frac{1}{2}+\frac{1}{12}+\frac{1}{8})\ &=\frac{6 - 4+3}{6}-(\frac{-12 + 2+3}{24})\ &=\frac{5}{6}-(-\frac{7}{24})\ &=\frac{20 + 7}{24}\ &=\frac{9}{8} \end{align*} ]

Answer:

C. $\frac{9}{8}$