MHT CET · Maths · Definite Integration
\(\int_0^2 \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{d} x=\)
- A \(\mathrm{a}-2 \mathrm{a} \log 2\)
- B \(a-a \log 2\)
- C \(\mathrm{a}+2 \mathrm{a} \log 2\)
- D \(a+a \log 2\)
Answer & Solution
Correct Answer
(A) \(\mathrm{a}-2 \mathrm{a} \log 2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Let } \mathrm{I}=\int_0^{\mathrm{a}} \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{~d} x \\
& \text { Let } x+\mathrm{a}=\mathrm{t} \\
& \Rightarrow x=\mathrm{t}-\mathrm{a} \\
& \text { If } x=0 \text {, then } \mathrm{t}=\mathrm{a} \\
& \text { If } x=\mathrm{a} \text {, then } \mathrm{t}=2 \mathrm{a} \\
& \therefore \quad \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=\int_a^{2 a} \frac{t-2 a}{t} d t \\
& =\int_a^{2 a} 1 d t-2 a \int_a^{2 a} \frac{1}{t} d t=[t]_a^{2 a}-2 a[\log t]_a^{2 a} \\
& =\mathrm{a}-2 \mathrm{a}(\log 2 \mathrm{a}-\log \mathrm{a}) \\
& =a-2 a \log 2
\end{aligned}\)
& \text { Let } \mathrm{I}=\int_0^{\mathrm{a}} \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{~d} x \\
& \text { Let } x+\mathrm{a}=\mathrm{t} \\
& \Rightarrow x=\mathrm{t}-\mathrm{a} \\
& \text { If } x=0 \text {, then } \mathrm{t}=\mathrm{a} \\
& \text { If } x=\mathrm{a} \text {, then } \mathrm{t}=2 \mathrm{a} \\
& \therefore \quad \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=\int_a^{2 a} \frac{t-2 a}{t} d t \\
& =\int_a^{2 a} 1 d t-2 a \int_a^{2 a} \frac{1}{t} d t=[t]_a^{2 a}-2 a[\log t]_a^{2 a} \\
& =\mathrm{a}-2 \mathrm{a}(\log 2 \mathrm{a}-\log \mathrm{a}) \\
& =a-2 a \log 2
\end{aligned}\)
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