Question Number 207938 by necx122 last updated on 31/May/24
$${what}\:{is}\:{the}\:{area}\:{bounded}\:{by}\:{the}\:{curve} \\ $$$${y}={x}\left({x}−\mathrm{2}\right)\left({x}−\mathrm{5}\right)\:{and}\:{the}\:{x}\:{axis}? \\ $$$$ \\ $$
Answered by mr W last updated on 31/May/24
![A=∫_0 ^2 x(x−2)(x−5)dx−∫_2 ^5 x(x−2)(x−5)dx =[(x^4 /4)−((7x^3 )/3)+5x^2 ]_0 ^2 −[(x^4 /4)−((7x^3 )/3)+5x^2 ]_2 ^5 =2×[(2^4 /4)−((7×2^3 )/3)+5×2^2 ]−[(5^4 /4)−((7×5^3 )/3)+5×5^2 ] =((253)/(12))](https://www.tinkutara.com/question/Q207941.png)
$${A}=\int_{\mathrm{0}} ^{\mathrm{2}} {x}\left({x}−\mathrm{2}\right)\left({x}−\mathrm{5}\right){dx}−\int_{\mathrm{2}} ^{\mathrm{5}} {x}\left({x}−\mathrm{2}\right)\left({x}−\mathrm{5}\right){dx} \\ $$$$=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{7}{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{5}{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{2}} −\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{7}{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{5}{x}^{\mathrm{2}} \right]_{\mathrm{2}} ^{\mathrm{5}} \\ $$$$=\mathrm{2}×\left[\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{7}×\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{5}×\mathrm{2}^{\mathrm{2}} \right]−\left[\frac{\mathrm{5}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{7}×\mathrm{5}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{5}×\mathrm{5}^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{253}}{\mathrm{12}} \\ $$