Question Number 32756 by NECx last updated on 01/Apr/18
$${Please}\:{help} \\ $$$$ \\ $$$${Find}\:{the}\:{area}\:{bounded}\:{by} \\ $$$${y}\left({x}+\mathrm{2}\right)={x}^{\mathrm{4}} ,{x}=\mathrm{0},{y}=\mathrm{0},{and}\:{x}=\mathrm{3} \\ $$
Answered by MJS last updated on 02/Apr/18
![y(x)=(x−2)^4 area between x−axis (y=0) and y(x) in [0;3], no change of sign ∫_0 ^3 (x−2)^4 dx=(1/5)(x−2)^5 ∣_0 ^3 = =(1/5)(1−(−2)^5 )=((33)/5)](https://www.tinkutara.com/question/Q32770.png)
$${y}\left({x}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{4}} \\ $$$$\mathrm{area}\:\mathrm{between}\:{x}−\mathrm{axis}\:\left({y}=\mathrm{0}\right)\:\mathrm{and} \\ $$$${y}\left({x}\right)\:\mathrm{in}\:\left[\mathrm{0};\mathrm{3}\right],\:\mathrm{no}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left({x}−\mathrm{2}\right)^{\mathrm{4}} {dx}=\frac{\mathrm{1}}{\mathrm{5}}\left({x}−\mathrm{2}\right)^{\mathrm{5}} \underset{\mathrm{0}} {\overset{\mathrm{3}} {\mid}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{1}−\left(−\mathrm{2}\right)^{\mathrm{5}} \right)=\frac{\mathrm{33}}{\mathrm{5}} \\ $$
Commented by NECx last updated on 02/Apr/18
$${please}\:{I}\:{dont}\:{understand}\:{how}\:{you} \\ $$$${got}\:{y}\left({x}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{4}} \\ $$$$…..{Theres}\:{nothing}\:{of}\:{such}\:{in}\:{the} \\ $$$${question}. \\ $$
Commented by MJS last updated on 02/Apr/18
![I understood that it′s y(x+2), not y×(x+2). If this is true, y(x+2)=x^4 ⇒ y(x)=(x−2)^4 [y(x)=(x−2)^4 with x=t+2= =y(t+2)=(t+2−2)^4 =t^4 ] if it′s y×(x+2)=x^4 ⇒ ⇒ y=(x^4 /(x+2))=x^3 −2x^2 +4x−8+((16)/(x+2)) still no change of sign in [0;3] ∫_0 ^3 x^3 −2x^2 +4x−8+((16)/(x+2))dx= =(1/4)x^4 −(4/3)x^3 +2x^2 −8x+16ln ∣x+2∣∣_0 ^3 = =−((15)/4)+16ln n−16ln 2= =−((15)/4)+16ln (5/2)≈10.91](https://www.tinkutara.com/question/Q32774.png)
$$\mathrm{I}\:\mathrm{understood}\:\mathrm{that}\:\mathrm{it}'\mathrm{s}\:{y}\left({x}+\mathrm{2}\right),\:\mathrm{not} \\ $$$${y}×\left({x}+\mathrm{2}\right).\:\mathrm{If}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true}, \\ $$$${y}\left({x}+\mathrm{2}\right)={x}^{\mathrm{4}} \:\Rightarrow\:{y}\left({x}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\left[{y}\left({x}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{4}} \:\mathrm{with}\:{x}={t}+\mathrm{2}=\right. \\ $$$$\left.\:\:\:\:\:\:\:={y}\left({t}+\mathrm{2}\right)=\left({t}+\mathrm{2}−\mathrm{2}\right)^{\mathrm{4}} ={t}^{\mathrm{4}} \right] \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{it}'\mathrm{s}\:{y}×\left({x}+\mathrm{2}\right)={x}^{\mathrm{4}} \:\Rightarrow \\ $$$$\Rightarrow\:{y}=\frac{{x}^{\mathrm{4}} }{{x}+\mathrm{2}}={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{8}+\frac{\mathrm{16}}{{x}+\mathrm{2}} \\ $$$$\mathrm{still}\:\mathrm{no}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{3}\right] \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{8}+\frac{\mathrm{16}}{{x}+\mathrm{2}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16ln}\:\mid{x}+\mathrm{2}\mid\underset{\mathrm{0}} {\overset{\mathrm{3}} {\mid}}= \\ $$$$=−\frac{\mathrm{15}}{\mathrm{4}}+\mathrm{16ln}\:{n}−\mathrm{16ln}\:\mathrm{2}= \\ $$$$=−\frac{\mathrm{15}}{\mathrm{4}}+\mathrm{16ln}\:\frac{\mathrm{5}}{\mathrm{2}}\approx\mathrm{10}.\mathrm{91} \\ $$
Commented by NECx last updated on 02/Apr/18
$${wow}….{now}\:{I}\:{understand} \\ $$$$ \\ $$