Blog: `int_0^1 (dx)/(1 + sqrt(x))^4` Evaluate the definite integral.

Wednesday, October 7, 2015

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({1}+\sqrt{{{x}}}\right)}}^{{4}}}$ Evaluate the definite integral.

You need to solve the definite integral, using fundamental theorem of calculus, such that:

$\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({b}\right)}-{F}{\left({a}\right)}$

First, you need to solve the indefinite integral $\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}$ , using the substitution $\displaystyle{1}+\sqrt{{x}}={t}$ such that:

$\displaystyle{1}+\sqrt{{x}}={t}\Rightarrow\frac{{1}}{{{2}\sqrt{{x}}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}={2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}$

$\displaystyle\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}=\int\frac{{{2}{t}}}{{{{t}}^{{4}}}}{\left.{d}{t}\right.}-\int\frac{{2}}{{{{t}}^{{4}}}}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}=\int\frac{{2}}{{{{t}}^{{3}}}}{\left.{d}{t}\right.}-\int\frac{{2}}{{{{t}}^{{4}}}}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}=\int{2}\cdot{\left({{t}}^{{-{3}}}\right)}{\left.{d}{t}\right.}-\int{2}\cdot{\left({{t}}^{{-{4}}}\right)}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}={2}\cdot\frac{{{{t}}^{{-{2}}}}}{{-{2}}}-{2}\frac{{{{t}}^{{-{3}}}}}{{-{3}}}+{c}$

$\displaystyle\int\frac{{{2}{\left({t}-{1}\right)}{\left.{d}{t}\right.}}}{{{{t}}^{{4}}}}=-\frac{{1}}{{{{t}}^{{2}}}}+\frac{{2}}{{{3}{{t}}^{{3}}}}+{c}$

Replacing back $\displaystyle{1}+\sqrt{{x}}$ for t yields:

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=-\frac{{1}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{2}}}}+\frac{{2}}{{{3}{{\left({1}+\sqrt{{x}}\right)}}^{{3}}}}+{c}$

Calculating the integral yields:

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}={\left(-\frac{{1}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{2}}}}+\frac{{2}}{{{3}{{\left({1}+\sqrt{{x}}\right)}}^{{3}}}}\right)}{{\mid}_{{0}}^{{1}}}$

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}={\left(-\frac{{1}}{{{{\left({1}+{1}\right)}}^{{2}}}}+\frac{{2}}{{{3}{{\left({1}+{1}\right)}}^{{3}}}}+\frac{{1}}{{{{\left({1}+{0}\right)}}^{{2}}}}-\frac{{2}}{{{3}{{\left({1}+{0}\right)}}^{{3}}}}\right)}$

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=-\frac{{1}}{{4}}+\frac{{1}}{{12}}+{1}-\frac{{2}}{{3}}$

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=\frac{{-{3}+{1}+{12}-{8}}}{{12}}$

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=\frac{{2}}{{12}}$

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=\frac{{1}}{{6}}$

Hence, evaluating the definite integral, using the fundamental theorem of calculus, yields $\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{{\left({1}+\sqrt{{x}}\right)}}^{{4}}}}=\frac{{1}}{{6}}.$

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Wednesday, October 7, 2015

$\displaystyle{\int_{{0}}^{{1}}}\frac{{{\left.{d}{x}\right.}}}{{{\left({1}+\sqrt{{{x}}}\right)}}^{{4}}}$ Evaluate the definite integral.

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