question_answer

Evaluate \[\int_{0}^{\pi /2}{\frac{{{\sin }^{4/5}}x}{{{\cos }^{4/5}}x+{{\sin }^{4/5}}x}\,dx.}\]

OR

Evaluate integrals as a limit of sum \[\int_{2}^{4}{{{2}^{x}}\,dx.}\]

Answer:

Let \[l=\int_{0}^{\pi /2}{\frac{{{\sin }^{4/5}}x}{{{\cos }^{4/5}}x+{{\sin }^{4/5}}x}\,dx}\]                  ?(i)

\[\Rightarrow \] \[l=\int_{0}^{\pi /2}{\frac{{{\sin }^{4/5}}\left( \frac{\pi }{2}-x \right)}{{{\cos }^{4/5}}\left( \frac{\pi }{2}-x \right)+{{\sin }^{4/5}}\left( \frac{\pi }{2}-x \right)}\,dx}\]

\[\left[ \because \,\,\,\int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]\]

\[\Rightarrow \]   \[l=\int_{0}^{\pi /2}{\frac{{{\cos }^{4/5}}x}{{{\sin }^{4/5}}x+{{\cos }^{4/5}}x}\,dx}\]              ?(ii)

\[\left[ \because \,\,\,\sin \left( \frac{\pi }{2}-x \right)=\cos x\,\,\,\text{and}\,\,\,\cos \left( \frac{\pi }{2}-x \right)=\sin x \right]\]On adding Eqs. (i) and (ii), we get

\[2l=\int_{0}^{\pi /2}{\frac{{{\sin }^{4/5}}x+{{\cos }^{4/5}}x}{{{\sin }^{4/5}}x+{{\cos }^{4/5}}x}\,dx}\]

\[\Rightarrow \]   \[2l=\int_{0}^{\pi /2}{1dx}\] \[\Rightarrow \] \[2l=[x]_{0}^{\pi /2}=\frac{\pi }{2}\] \[\Rightarrow \]

\[l=\frac{\pi }{4}\]

Hence, \[\int_{0}^{\pi /2}{\frac{{{\sin }^{4/5}}x}{{{\cos }^{4/5}}x+{{\sin }^{4/5}}x}\,dx}=\frac{\pi }{4}\]

OR

Given integral is \[\int_{2}^{4}{{{2}^{x}}\,dx.}\]

We know that,

\[\int_{a}^{b}{f(x)dx}=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,h[f(a)+f(a+h)+f(a+2h)\]

\[+...+f\{a+(n-1)h\}]\]

Where, \[h=\frac{b-a}{n}\]

Here a = 2, b = 4, \[f(x)={{2}^{x}}\] and \[h=\frac{4-2}{n}\]

\[\Rightarrow \] nh = 2

\[\therefore \] \[\int_{2}^{4}{{{2}^{x}}dx=\underset{h\to \,\,0}{\mathop{\lim }}\,}\,\,h[f(2)+f(2+h)+f(2+2h)\]

\[+...+f\{2+(n-1)h\}\]

\[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,h[{{2}^{2}}+{{2}^{2\,+\,h}}+{{2}^{2\,+\,2h}}+...+{{2}^{2+(n\,-\,1)h}}]\]            \[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,4h[1+{{2}^{h}}+{{2}^{2h}}+...+{{2}^{(n-1)h}}]\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\,\,4h\left[ \frac{{{({{2}^{h}})}^{n}}-1}{{{2}^{h}}-1} \right]\]

[\[\because \] sum of first and terms of GP is

\[\left. {{S}_{n}}=a\left( \frac{{{r}^{n}}-1}{r-1} \right),\,r\ne 1 \right]\]

\[=4\,\,\underset{h\to \,\,0}{\mathop{\lim }}\,\left[ \frac{{{2}^{nh}}-1}{\left( \frac{{{2}^{h}}-1}{h} \right)} \right]\]

\[=4\times \left( \frac{{{2}^{2}}-1}{\log \,\,2} \right)\] \[\left[ \because nh=2\,\,\,\text{and}\,\,\,\underset{h\to \,\,0}{\mathop{\lim }}\,\frac{{{2}^{h}}-1}{h}=\log 2 \right]\]

\[=\frac{12}{\log \,\,2}\]