Answer to Question #110233 in Calculus for lolipop

Question

[img]https://upload.cc/i1/2020/04/16/pcQUg6.jpg[/img]


question in photo
R=4

Answer 1

a) $f(x)=\int _{(R+1)^3} ^{x^3} \sqrt {1+e^{2t}} dt$

i) Using Newton-Leibnitz's rule; we get;

$f'(x)=3x^2\sqrt { 1+e^{2(x^3)}}- 0$

$\implies f'(x)=3x^2\sqrt { 1+e^{2x^3}}$ ----(1) ---(Answer)

ii) Given : $g(x)=sin(f(x)) \implies g'(x)=cos(f(x))*f'(x)$ ---(2)

$\implies g'(R+1)=cos(f(R+1))*f'(R+1)$

Now, $f(R+1)=\int _{(R+1)^3} ^{(R+1)^3} \sqrt {1+e^{2t}} dt=0$ ---(3)

$(1) \implies f'(R+1)=3(R+1)^2\sqrt { 1+e^{2(R+1)^3}}$ ---(4)

Using (3) and (4) in (2), we get;

$g'(R+1)=cos(0)*f'(R+1)=f'(R+1)$

$\implies g'(R+1)=3(R+1)^2\sqrt { 1+e^{2(R+1)^3}}$

Also given; R=4 \implies$\implies g'(5)=3(5)^2\sqrt { 1+e^{2(5)^3}}=375\sqrt { 1+e^{250}}$

$=7.25*10^{56}$

b) Given : n is a positive integer.

i) Let $I_n=\int _0 ^{\pi /4}tan^nx.dx$

Then, $I_{n+1}=\int _0 ^{\pi /4}tan^{n+1}x.dx$

$=\int _0 ^{\pi /4}tanx*tan^nxdx$

$=[tanx]_0 ^{\pi /4}*\int _0 ^{\pi /4}tan^nxdx-\int _0 ^{\pi /4}sec^2x*(\int _0 ^{\pi /4}tan^nxdx)dx$

$=(1-0)I_n-\int _0 ^{\pi /4}sec^2x*I_ndx$

$\implies I_{n+1}=I_n-\int _0 ^{\pi /4}sec^2x*I_ndx$ ---(1)

As n is a positive integer and $tanx$ is positive in the interval $(0,\pi /4) \implies I_n>0$

Also, $sec^2x >0 \implies \int _0 ^{\pi /4}sec^2x*I_ndx >0$ ---(2)

Thus, $(1);(2) \implies I_{n+1}<I_n$

ii) For this, we use the definition that Integration of a function is area under the curve of that function.

Let $f(x)=x^n; g(x)=x^n/(1+x^3)$

Given : $0<x<1 \implies 1+x^3>1 \implies 1/(1+x^3)<1$

$\implies x^n/(1+x^3)<x^n$

$\implies f(x)>g(x); \forall x \in (0,1)$

Using the above information we sketch approximate graphs of the two functions:

Clearly, $\int _0 ^1f(x)dx>\int _0 ^1g(x)dx; \forall x \in (0,1)$ (as area under g(x) is smaller)

$\implies \int _0 ^1x^n/(1+x^3)dx<\int _0 ^1x^n dx$ -----(Answer)