Question

Evaluate the definite integral

Short answer

Hence the \(\int_{\rm{0}}^{\rm{4}} {\frac{{\rm{x}}}{{\sqrt {{\rm{1 + 2x}}} }}} {\rm{dx}}\) is equal to \(\frac{{{\rm{10}}}}{{\rm{3}}}\)is proved

Step-by-step solution

Find \({\rm{u}}\)and\({\rm{du}}\).

Finding \({\rm{u}}\) and\({\rm{du}}\) value is \({\rm{u = 1 + 2x }} \Rightarrow {\rm{x = }}\frac{{{\rm{u - 1}}}}{{\rm{2}}}\)

\({\rm{du = 2dx }} \Rightarrow \frac{{{\rm{du}}}}{{\rm{2}}}{\rm{ = dx}}\)

The new integration constraints for\({\rm{u}}\).

\({\rm{u = 9, if x = 4}}\)

\({\rm{u = 1 , if x = 0}}\)

Substituting \({\rm{u}}\) in order to work\(\int_{\rm{0}}^{\rm{4}} {\frac{{\rm{x}}}{{\sqrt {{\rm{1 + 2x}}} }}} {\rm{dx}}\).

\(\begin{aligned}{c}\frac{{\rm{x}}}{{\rm{0}}}\frac{{\rm{x}}}{{\sqrt {{\rm{1 + 2x}}} }}{\rm{dx = }}\frac{{\rm{9}}}{{\frac{{{\rm{u - 1}}}}{{\rm{2}}}}}\frac{{{\rm{du}}}}{{\sqrt {\rm{u}} }}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\;\;\;{{\rm{u}}^{{\rm{ - 1/2}}}}{\rm{(u - 1)du}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\;\;\;\frac{{\rm{9}}}{{\rm{9}}}\;\;\;\left( {{{\rm{u}}^{{\rm{1/2}}}}{\rm{ - }}{{\rm{u}}^{{\rm{ - 1/2}}}}} \right){\rm{du}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\frac{{{{\rm{u}}^{{\rm{3/2}}}}}}{{{\rm{3/2}}}}{\rm{ - }}\left. {\frac{{{{\rm{u}}^{{\rm{1/2}}}}}}{{{\rm{1/2}}}}} \right|_{\rm{1}}^{\rm{9}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\frac{{{{\rm{9}}^{{\rm{3/2}}}}}}{{{\rm{3/2}}}}{\rm{ - }}\frac{{{{\rm{9}}^{{\rm{1/2}}}}}}{{{\rm{1/2}}}}{\rm{ - }}\frac{{{{\rm{1}}^{{\rm{3/2}}}}}}{{{\rm{3/2}}}}{\rm{ - }}\frac{{{{\rm{1}}^{{\rm{1/2}}}}}}{{{\rm{1/2}}}}\end{aligned}\)

\({\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times 12 - }}\left( {{\rm{ - }}\frac{{\rm{4}}}{{\rm{3}}}} \right)\)

\(\begin{aligned}{c}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\frac{{{\rm{40}}}}{{\rm{3}}}\\{\rm{ = }}\frac{{{\rm{10}}}}{{\rm{3}}}\end{aligned}\)

Hence the \(\int_{\rm{0}}^{\rm{4}} {\frac{{\rm{x}}}{{\sqrt {{\rm{1 + 2x}}} }}} {\rm{dx}}\) is equal to \(\frac{{{\rm{10}}}}{{\rm{3}}}\)is proved.

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