Question 4c(ii)

SCH_CALC L3CALC

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$$I={\int }_a^{a+1}\sin (e^x)dx$$

Let the antiderivative of $\sin e^x=F(x)$ because $\int \sin e^{x}dx\ne -\frac{\cos e^x}{e^x}+C$

$$I=F(a+1)-F(a)$$ $$\frac{dI}{da}=F^{\prime }(a+1)-F^{\prime }(a)$$ $$\frac{dI}{da}=\sin (e^{a+1})-\sin (e^a)=0$$ $$\sin (e^{a+1})=\sin e^a$$ $$⟹e^{a+1}=e^a$$

but the above equation is false — there are no values for $a$ that satisfies the above equation!

Use general solution of $\sin x$!

$$\alpha ={\sin }^{-1}(\sin (e^a))\equiv e^a$$ $$e^{a+1}=n\pi +(-1{)}^n\cdot e^a$$ $$e^a\cdot e^1-(-1{)}^n\cdot e^a=n\pi$$ $$e^a(e-(-1{)}^n)=n\pi$$ $$e^a=\frac{n\pi }{e-(-1{)}^n}$$ $$a=\ln \left(\frac{n\pi }{e-(-1{)}^n}\right)$$

if $n=0$:

$$⟹a=\ln \left(\frac{0}{e-(-1{)}^0}\right)\equiv 0$$

if $n=1$:

$$⟹a=\ln \left(\frac{\pi }{e+1}\right)\approx -0.1685$$

if $n=2$:

$$⟹a=\ln \left(\frac{2\pi }{e-(-1{)}^2}\right)\approx 1.2966$$

if $n=3$:

$$⟹a=\ln \left(\frac{3\pi }{e-(-1{)}^3}\right)\approx 0.9301$$

if $n=4$:

$$⟹a=\ln \left(\frac{4\pi }{e-(-1{)}^4}\right)\approx 1.9897$$

Looking at the graph, the maximum area is generated when $n=1,a=-0.1685$, since when $n=2,3,4\dots$ the area gets smaller..

In exact form $a=\ln \left(\frac{\pi }{e+1}\right)$!