Tip:A–D to answerE for explanationV for videoS to reveal answer
Let the area of the region {(x,y):0≤x≤3,0≤y≤min(x2+2,2x+2)} be A. Then 12A is equal to .........
- A.
164
(Correct Answer) - B.
146
- C.
165
- D.
None
Explanation
Region = {(x, y): 0 ≤ x ≤ 3, 0 < y < min(x2 + 2, 2x + 2)}
\begin{aligned}
\mathrm{A} & =\int_{0}^{2}x^{2}+2+\int_{2}^{3}2x+2 \\
& =\left[\frac{x^{3}}{3}+2x\right]_{0}^{2}+[x^{2}+2x]_{2}^{3} \\
& =\frac{8}{3}+4+\left[9+6-4-4\right] \\
& =\frac{8}{3}+4+7 \\
& =11+\frac{8}{3} \\
\mathrm{12~A} & =12\times11+32 \\
& =132+32 \\
& =164
\end{aligned}
Explanation
Region = {(x, y): 0 ≤ x ≤ 3, 0 < y < min(x2 + 2, 2x + 2)}
\begin{aligned}
\mathrm{A} & =\int_{0}^{2}x^{2}+2+\int_{2}^{3}2x+2 \\
& =\left[\frac{x^{3}}{3}+2x\right]_{0}^{2}+[x^{2}+2x]_{2}^{3} \\
& =\frac{8}{3}+4+\left[9+6-4-4\right] \\
& =\frac{8}{3}+4+7 \\
& =11+\frac{8}{3} \\
\mathrm{12~A} & =12\times11+32 \\
& =132+32 \\
& =164
\end{aligned}
