Answered: "Please answer with solution."

asistioacer

Answered question

2022-02-24

Please answer with solution.

alenahelenash

Expert2022-03-10Added 556 answers

2. Since $y = x^{2}$ and $y = 4 x - x^{2}$

$x^{2} = 4 x - x^{2}$

$2 x^{2} - 4 x = 0 \Rightarrow 2 x \cdot (x - 2) = 0$

$2 x = 0 \Rightarrow x = 0$

since $x = 0 \Rightarrow y = x^{2} \Rightarrow y = 0$

$A = (0, 0)$

$x - 2 = 0 \Rightarrow x = 2$

since $x = 2 \Rightarrow y = x^{2} \Rightarrow y = 4$

$B = (2, 4)$

if we take $d x$ slice (that mean the Area revolving about x-axis)

The interval of the integral become

$x \in [0, 2]$

now let set up the integral of Area

$A = \int_{a}^{b} y_{2} - y_{1} * d x$

$A = \int_{0}^{2} (4 x - x^{2}) - (x^{2}) \cdot d x$

$A = \int_{0}^{2} 4 x - 2 x^{2} \cdot d x = {[2 x^{2} - \frac{2}{3} x^{3}]}_{0}^{2}$

$[8 - \frac{16}{3}] = \frac{24 - 16}{3} = \frac{8}{3} = 2.667 {(u n i t e)}^{2}$

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