English

If F' (X) = X − 1 X 2 1 2 , - Mathematics

Advertisements
Advertisements

Question

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 

Sum

Solution

\[f'\left( x \right) = x - \frac{1}{x^2}\]
\[ f'\left( x \right) = x - x^{- 2} \]
\[\int f'\left( x \right)dx = \int\left( x - x^{- 2} \right)dx\]
\[ f\left( x \right) = \frac{x^2}{2} - \frac{x^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \frac{x^2}{2} + \frac{1}{x} + C\]
\[f\left( 1 \right) = \frac{1}{2} \left( Given \right)\]
\[ \Rightarrow \frac{1^2}{2} + \frac{1}{1} + C = \frac{1}{2}\]
\[ \Rightarrow C = - 1\]
\[ \therefore f\left( x \right) = \frac{x^2}{2} + \frac{1}{x} - 1\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 15]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 45 | Page 15

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]

\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

\[\int\frac{1 - \sin 2x}{x + \cos^2 x} dx\]

\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]

` ∫   e^{m   sin ^-1  x}/ \sqrt{1-x^2}  ` dx

 


\[\int\frac{e^{2x}}{1 + e^x} dx\]

\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]

` ∫  sec^6   x  tan    x   dx `

\[\int \cot^5 x  \text{ dx }\]

Evaluate the following integrals:

\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]

\[\int\frac{1}{x^2 + 6x + 13} dx\]

\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]

\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]

`int 1/(cos x - sin x)dx`

\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]

\[\int x^2 \cos 2x\ \text{ dx }\]

\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]

\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]

\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]

\[\int\frac{1}{\sin x \left( 3 + 2 \cos x \right)} dx\]

\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{  dx }\]

\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]

\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

\[\int\frac{1}{e^x + 1} \text{ dx }\]

\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]

\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]


\[\int\frac{x}{x^3 - 1} \text{ dx}\]

\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]

\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]


\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]


Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×