Question Number 116687 by megrex last updated on 05/Oct/20
![Help please, to solve this ... If f(x)=1+x^2 for x∈[−2,2] and f(x)=5 otherwise. Then what is the value of ∫_(−2) ^(+2) f(2x^2 )dx?](Q116687.png)
$${Help}\:{please},\:{to}\:{solve}\:{this}\:... \\ $$$${If}\:{f}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{2}} \:\:{for}\:{x}\in\left[−\mathrm{2},\mathrm{2}\right]\:{and}\: \\ $$$$\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{5}\:\:\:\:\:\:\:\:{otherwise}. \\ $$$${Then}\:{what}\:{is}\:{the}\:{value}\:{of} \\ $$$$\int_{−\mathrm{2}} ^{+\mathrm{2}} {f}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx}? \\ $$$$ \\ $$
Answered by Olaf last updated on 05/Oct/20
$${f}\:\mathrm{is}\:\mathrm{even}\:\mathrm{and}\:\mathrm{I}\:=\:\int_{−\mathrm{2}} ^{+\mathrm{2}} {f}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:\leqslant\:\mathrm{2}\:\Leftrightarrow\:{x}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:{f}\left(\mathrm{2}{x}^{\mathrm{2}} \right)\:=\:\mathrm{1}+\mathrm{4}{x}^{\mathrm{4}} \\ $$$${f}\left(\mathrm{2}{x}^{\mathrm{2}} \right)\:=\:\mathrm{5}\:\mathrm{otherwise}. \\ $$$$ \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{4}} \right){dx}+\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{5}{dx} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{5}}\right)+\mathrm{10}\:=\:\frac{\mathrm{68}}{\mathrm{5}} \\ $$
Commented by megrex last updated on 05/Oct/20
$${Thankyou}\:{Olaf}. \\ $$
Answered by Olaf last updated on 05/Oct/20
![other method : u = 2x^2 du = 4xdx = 4(√(u/2))dx dx = (du/( 2(√2)(√u))) I = 2∫_0 ^(+2) f(2x^2 )dx = (1/( (√2)))∫_0 ^8 f(u)(du/( (√u))) I = (1/( (√2)))∫_0 ^2 ((1+u^2 )/( (√u)))du+(1/( (√2)))∫_2 ^8 (5/( (√u)))du I = (1/( (√2)))[2(√u)+(2/5)u^(5/2) ]_0 ^2 +(1/( (√2)))[10(√u)]_2 ^8 I = 2+(8/5)+(1/( (√2)))(10(√8)−10(√2)) I = 2+(8/5)+10 = ((68)/5)](Q116690.png)
$$ \\ $$$$\mathrm{other}\:\mathrm{method}\:: \\ $$$${u}\:=\:\mathrm{2}{x}^{\mathrm{2}} \\ $$$${du}\:=\:\mathrm{4}{xdx}\:=\:\mathrm{4}\sqrt{\frac{{u}}{\mathrm{2}}}{dx} \\ $$$${dx}\:=\:\frac{{du}}{\:\mathrm{2}\sqrt{\mathrm{2}}\sqrt{{u}}} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{+\mathrm{2}} {f}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{8}} {f}\left({u}\right)\frac{{du}}{\:\sqrt{{u}}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}+{u}^{\mathrm{2}} }{\:\sqrt{{u}}}{du}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{2}} ^{\mathrm{8}} \frac{\mathrm{5}}{\:\sqrt{{u}}}{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{2}\sqrt{{u}}+\frac{\mathrm{2}}{\mathrm{5}}{u}^{\mathrm{5}/\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{10}\sqrt{{u}}\right]_{\mathrm{2}} ^{\mathrm{8}} \\ $$$$\mathrm{I}\:=\:\mathrm{2}+\frac{\mathrm{8}}{\mathrm{5}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{10}\sqrt{\mathrm{8}}−\mathrm{10}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{I}\:=\:\mathrm{2}+\frac{\mathrm{8}}{\mathrm{5}}+\mathrm{10}\:=\:\frac{\mathrm{68}}{\mathrm{5}} \\ $$
Commented by megrex last updated on 05/Oct/20
$${Very}\:{nice}!\:\:{Many}\:{thanks}. \\ $$