Question Number 60670 by Meritguide1234 last updated on 24/May/19
Commented by Meritguide1234 last updated on 24/May/19
[.]=greatest integer Function
Commented by rahul 19 last updated on 24/May/19
$$\mathrm{8}.\mathrm{000000} \\ $$
Answered by tanmay last updated on 24/May/19
![∫_0 ^π e^x (((1−cos2x)/2))dx (1/2)∫_0 ^π e^x dx−(1/2)∫_0 ^π e^x cos2xdx I_1 =((e^π −1)/2) I_2 =(1/2)∫_0 ^π e^x cos2xdx I=I_1 −I_2 formula ∫e^(ax) cosbxdx=((e^(ax) (acosbx+bsinbx))/(a^2 +b^2 )) here a=1 b=2 =(1/2)[((e^x (cos2x+2sin2x))/5)] so I_2 =(1/2)∣((e^x (cos2x+2sin2x))/5)∣_0 ^π I_2 =(1/(10))[e^π (cos2π+2sin2π)−e^0 (cos0+2sin0)] =(1/(10))[e^π (1)−1] I=I_1 −I_2 =((e^π −1)/2)−((e^π −1)/(10)) =((5e^π −5−e^π +1)/(10)) =((4e^π −4)/(10)) =(2/5)(e^π −1) ≈8.86 so required answer is ⌊8.86⌋ =8](https://www.tinkutara.com/question/Q60674.png)
$$\int_{\mathrm{0}} ^{\pi} {e}^{{x}} \left(\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {cos}\mathrm{2}{xdx} \\ $$$${I}_{\mathrm{1}} =\frac{{e}^{\pi} −\mathrm{1}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {cos}\mathrm{2}{xdx} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${formula}\:\int{e}^{{ax}} {cosbxdx}=\frac{{e}^{{ax}} \left({acosbx}+{bsinbx}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${here}\:{a}=\mathrm{1}\:\:\:{b}=\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{2}{sin}\mathrm{2}{x}\right)}{\mathrm{5}}\right] \\ $$$${so}\:{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{2}{sin}\mathrm{2}{x}\right)}{\mathrm{5}}\mid_{\mathrm{0}} ^{\pi} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{10}}\left[{e}^{\pi} \left({cos}\mathrm{2}\pi+\mathrm{2}{sin}\mathrm{2}\pi\right)−{e}^{\mathrm{0}} \left({cos}\mathrm{0}+\mathrm{2}{sin}\mathrm{0}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left[{e}^{\pi} \left(\mathrm{1}\right)−\mathrm{1}\right] \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} =\frac{{e}^{\pi} −\mathrm{1}}{\mathrm{2}}−\frac{{e}^{\pi} −\mathrm{1}}{\mathrm{10}} \\ $$$$=\frac{\mathrm{5}{e}^{\pi} −\mathrm{5}−{e}^{\pi} +\mathrm{1}}{\mathrm{10}} \\ $$$$=\frac{\mathrm{4}{e}^{\pi} −\mathrm{4}}{\mathrm{10}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left({e}^{\pi} −\mathrm{1}\right) \\ $$$$\approx\mathrm{8}.\mathrm{86} \\ $$$${so}\:{required}\:{answer}\:{is}\:\lfloor\mathrm{8}.\mathrm{86}\rfloor \\ $$$$=\mathrm{8} \\ $$$$ \\ $$