Indefinite Integration 1 Question 58
59. If for non-zero $x, a f(x)+b f \quad \frac{1}{x}=\frac{1}{x}-5$, where $a \neq b$, then $\int _1^{2} f(x) d x=\ldots \ldots$.
(1996, 2M)
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Solution:
- Given, $a f(x)+b f(1 / x)=\frac{1}{x}-5$
Replacing $x$ by $1 / x$ in Eq. (i), we get
$$ a f(1 / x)+b f(x)=x-5 $$
On multiplying Eq. (i) by $a$ and Eq. (ii) by $b$, we get
$$ \begin{aligned} & a^{2} f(x)+a b f(1 / x)=a \frac{1}{x}-5 \\ & a b f(1 / x)+b^{2} f(x)=b(x-5) \end{aligned} $$
On subtracting Eq. (iv) from Eq. (iii), we get
$$ \begin{aligned} \left(a^{2}-b^{2}\right) f(x) & =\frac{a}{x}-b x-5 a+5 b \\ \Rightarrow \quad f(x) & =\frac{1}{\left(a^{2}-b^{2}\right)} \frac{a}{x}-b x-5 a+5 b \\ \Rightarrow \quad \int _1^{2} f(x) d x & =\frac{1}{\left(a^{2}-b^{2}\right)} \int _1^{2} \frac{a}{x}-b x-5 a+5 b \quad d x \\ & =\frac{1}{\left(a^{2}-b^{2}\right)} a \log |x|-\frac{b}{2} x^{2}-5(a-b) x{ } _1^{2} \\ & =\frac{1}{\left(a^{2}-b^{2}\right)}[a \log 2-2 b-10(a-b) \\ & =\frac{1}{\left(a^{2}-b^{2}\right)} a \log 2-5 a+\frac{7}{2} b \end{aligned} $$
