SATHEE: Indefinite Integration 1 Question 58

Indefinite Integration 1 Question 58

59. If for non-zero $x, a f (x) + b f \frac{1}{x} = \frac{1}{x} - 5$ , where $a \neq b$ , then $\int_{1}^{2} f (x) d x = \dots \dots$ .

(1996, 2M)

Show Answer

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

$\begin{aligned} (a^{2} - b^{2}) f (x) & = \frac{a}{x} - b x - 5 a + 5 b \\ \Rightarrow f (x) & = \frac{1}{(a^{2} - b^{2})} \frac{a}{x} - b x - 5 a + 5 b \\ \Rightarrow \int_{1}^{2} f (x) d x & = \frac{1}{(a^{2} - b^{2})} \int_{1}^{2} \frac{a}{x} - b x - 5 a + 5 b d x \\ = \frac{1}{(a^{2} - b^{2})} a \log | x | - \frac{b}{2} x^{2} - 5 (a - b) x_{1}^{2} \\ = \frac{1}{(a^{2} - b^{2})} [a \log 2 - 2 b - 10 (a - b) \\ = \frac{1}{(a^{2} - b^{2})} a \log 2 - 5 a + \frac{7}{2} b \end{aligned}$

Indefinite Integration 1 Question 58

59. If for non-zero $x, a f (x) + b f \frac{1}{x} = \frac{1}{x} - 5$ , where $a \neq b$ , then $\int_{1}^{2} f (x) d x = \dots \dots$ .

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Indefinite Integration 1 Question 58

59. If for non-zero x,af(x)+bf1x=1x−5, where a≠b, then ∫12f(x)dx=…….

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

59. If for non-zero $x, a f (x) + b f \frac{1}{x} = \frac{1}{x} - 5$ , where $a \neq b$ , then $\int_{1}^{2} f (x) d x = \dots \dots$ .