Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 58940 by Tony Lin last updated on 01/May/19

$$e^{i{\int }_{-2}^2(x^{2}sinx+\sqrt{1-\frac{x^2}{4}})dx}+\underset{x\to 2}{\lim }\frac{{\int }_2^{x}log(x+8)dx}{x-2}=?$$

Commented by maxmathsup by imad last updated on 01/May/19

$$letA={\int }_{-2}^2(x^{2}sinx+\sqrt{1-\frac{x^2}{4}})dx\Rightarrow A={\int }_{-2}^2{x}^{2}sinxdx+{\int }_{-2}^2\sqrt{1-\frac{x^2}{4}}dx$$$$x\to x^{2}sin\left(x\right)isoddfunction\Rightarrow {\int }_{-2}^2{x}^{2}sin\left(x\right)dx=0$$$${\int }_{-2}^2\sqrt{1-\frac{x^2}{4}}dx=2{\int }_0^2\sqrt{1-\frac{x^2}{4}}dx{=}_{\frac{x}{2}=sin\left(t\right)}2{\int }_0^{\frac{\pi }{2}}\sqrt{1-{sin}^{2}t}\left(2cost\right)dt$$$$=4{\int }_0^{\frac{\pi }{2}}{cos}^{2}tdt=4{\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2t\right)}{2}dt=2{\int }_0^{\frac{\pi }{2}}(1+cos\left(2t\right))dt$$$$=\pi +{\left[sin\left(2t\right)\right]}_0^{\frac{\pi }{2}}=\pi \Rightarrow e^{i{\int }_{-2}^2(x^{2}sinx+\sqrt{1-\frac{x^2}{4}})dx}=e^{i\pi }=-1$$$${\int }_2^{x}ln(t+8)dt{=}_{t+8=u}{\int }_{10}^{x+8}ln\left(u\right)du={[uln\left(u\right)-u]}_{10}^{x+8}$$$$=(x+8)ln(x+8)-(x+8)-10ln\left(10\right)+10\Rightarrow$$$${lim}_{x\to 2}\frac{{\int }_2^{x}ln(x+8)dx}{x-2}={lim}_{x\to 2}\frac{(x+8)ln(x+8)-(x+8)-10ln\left(10\right)+10}{x-2}$$$$={lim}_{x\to 2}\frac{u^{{}^{\prime }}\left(x\right)}{v^{{}^{\prime }}\left(x\right)}\left(usinghospitaltheorem\right)$$$$u\left(x\right)=(x+8)ln(x+8)-(x+8)-10ln\left(10\right)+10\Rightarrow$$$$u^{{}^{\prime }}\left(x\right)=ln(x+8)+1-1=ln(x+8)\Rightarrow u^{{}^{\prime }}\left(2\right)=ln\left(10\right)$$$$v\left(x\right)=x-2\Rightarrow v^{{}^{\prime }}\left(x\right)=1\Rightarrow {lim}_{x\to 2}\frac{{\int }_2^{x}ln(x+8)}{x-2}dx=ln\left(10\right)\Rightarrow$$$$e^{i{\int }_{-2}^2(....)}dx+{lim}_{x\to 2}\frac{{\int }_2^x(....)dx}{x-2}=-1+ln\left(10\right)$$$$hereihavetakenlnnotlogbecauselog\left(x\right)=\frac{ln\left(x\right)}{ln\left(10\right)}....$$$$$$

Commented by maxmathsup by imad last updated on 01/May/19

$$ifwetakelog(ddcimallog.)theresultwillbe-1+1=0$$

Commented by Tony Lin last updated on 02/May/19

$$thanks,sirXD$$

Answered by tanmay last updated on 02/May/19

$$log(x+8)=\frac{dF}{dx}$$$${\int }_2^x\frac{dF}{dx}\times dx$$$$=F\left(x\right)-F\left(2\right)$$$$now\underset{x\to 2}{\lim }\frac{F\left(x\right)-F\left(2\right)}{x-2}$$$$h=x-2$$$$\underset{h\to 0}{\lim }\frac{F(h+2)-F\left(2\right)}{h}$$$$=F^{{}^{\prime }}\left(2\right)$$$$=log(2+8)=log10$$

Commented by Tony Lin last updated on 02/May/19

$$welldone,sir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com