Science:Math Exam Resources/Courses/MATH101/April 2018/Question 10 (iii)

MATH101 April 2018

• Q1 (i) • Q1 (ii) • Q1 (iii) • Q1 (iv) • Q2 (i) • Q2 (ii) • Q2 (iii) • Q3 (i) • Q3 (ii) • Q4 (i) • Q4 (ii) • Q5 (i) • Q5 (ii) • Q6 (i) • Q6 (ii) • Q7 (i) • Q7 (ii) • Q8 (i) • Q8 (ii) • Q9 (i) • Q9 (ii) • Q10 (i) • Q10 (ii) • Q10 (iii) • Q11 (i) • Q11 (ii) •

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Question 10 (iii)
Now calculate the integral $I$ for the value of $c$ found in part 10 (ii).

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Hint
Use the integration by parts.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since $c$ is given by $2$ in part (ii), it is enough to evaluate the integral $\int _{0}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx.$ We apply the integration by parts for $u(x)=2x-\log(1+2x)$ and $v'(x)={\frac {1}{x^{2}}}$ . Then, $u'(x)=2-{\frac {2}{1+2x}}={\frac {4x}{1+2x}}$ , $v(x)=-{\frac {1}{x}}$ , and ${\begin{aligned}\int _{0}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx&=\lim _{t\to 0}\int _{t}^{1}{\frac {\left[2x-\log(1+2x)\right]}{x^{2}}}dx\\&=\lim _{t\to 0}\int _{t}^{1}u(x)v'(x)dx=\lim _{t\to 0}u(x)v(x){\bigg \|}_{t}^{1}-\int _{0}^{1}u'(x)v(x)dx\\&=\lim _{t\to 0}(2x-\log(1+2x))\cdot \left(-{\frac {1}{x}}\right){\bigg \|}_{t}^{1}-\int _{0}^{1}{\frac {4x}{1+2x}}\cdot \left(-{\frac {1}{x}}\right)dx\\&=-2+\log 3+\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}+\int _{0}^{1}{\frac {4}{1+2x}}dx.\end{aligned}}$ By L'hospital rule, we can compute the limit as $\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}=\lim _{t\to 0}{\frac {(2t-\log(1+2t))'}{(t)'}}=\lim _{t\to 0}{\frac {2-{\frac {2}{1+2t}}}{1}}=0.$ On the other hand, using the substitution $u=1+2x$ (so $du=2dx$ ), the last term can be evaluated; $\int _{0}^{1}{\frac {4}{1+2x}}dx=\int _{1}^{3}{\frac {2}{u}}du=2\log u{\bigg \|}_{1}^{3}=2\log 3.$ Combining all the information, the integral $I$ for $c=2$ has the value $I=-2+\log 3+\lim _{t\to 0}{\frac {(2t-\log(1+2t))}{t}}+\int _{0}^{1}{\frac {4}{1+2x}}dx=-2+\log 3+0+2\log 3=\log(27)-2.$ Answer: $\color {blue}\log(27)-2$