Use the mid-ordinate rule to estimate the area enclosed by the curve y = -x² + 2x + 8, the y-axis and x-axis using four strips.

Kwanza, badilisha nambari za desimali katika denominator kuwa sehemu:

$$2.125 = \frac{2125}{1000}$$

Gawanya nambari na denominator kwa 125:

$$2.125 = \frac{2125 \div 125}{1000 \div 125} = \frac{17}{8}$$

Pili, badilisha nambari ya pili ya desimali kuwa sehemu:

$$0.064 = \frac{64}{1000}$$

Sasa zidisha sehemu hizi mbili:

$$2.125 \times 0.064 = \frac{17}{8} \times \frac{64}{1000}$$

Futa 8 kutoka kwa 64:

$$= \frac{17 \times (8 \times 8)}{8 \times 1000} = \frac{17 \times 8}{1000}$$ $$= \frac{136}{1000}$$

Sasa weka denominator iliyorahisishwa kwenye usemi wa asili:

$$\sqrt[3]{\frac{93}{\frac{136}{1000}}}$$

Ili kugawanya kwa sehemu, zidisha kwa kinyume chake:

$$= \sqrt[3]{\frac{93 \times 1000}{136}}$$ $$= \sqrt[3]{\frac{93000}{136}}$$

Gawanya nambari na denominator kwa kigawanyo chao kikubwa zaidi. Tunaweza kugawanya zote mbili kwa 8:

$$93000 = 93 \times 1000 = 93 \times 125 \times 8$$ $$136 = 17 \times 8$$

Hivyo, sehemu inakuwa:

$$\frac{93 \times 125 \times 8}{17 \times 8} = \frac{93 \times 125}{17}$$

Sasa usemi ni:

$$\sqrt[3]{\frac{93 \times 125}{17}}$$

Tunajua kwamba $125 = 5^3$, kwa hivyo $\sqrt[3]{125} = 5$.

$$= \sqrt[3]{\frac{93}{17} \times 125}$$ $$= \sqrt[3]{\frac{93}{17}} \times \sqrt[3]{125}$$ $$= \sqrt[3]{\frac{93}{17}} \times 5$$ $$= 5 \sqrt[3]{\frac{93}{17}}$$

Thamani ya usemi ni:

$$5 \sqrt[3]{\frac{93}{17}}$$