हिंदी

If eeeed∫3ex-5e-x4e6x+5e-xdx = ax + b log |4ex + 5e –x| + C, then ______. - Mathematics

Advertisements
Advertisements

प्रश्न

If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.

विकल्प

  • a = `(-1)/8`, b = `7/8`

  • a = `1/8`, b = `7/8`

  • a = `(-1)/8`, b = `(-7)/8`

  • a = `1/8`, b = `(-7)/8`

MCQ
रिक्त स्थान भरें

उत्तर

If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then a = `(-1)/8`, b = `(-7)/8`.

Explanation:

`(3"e"^x - 5"e"^-x)/(4"e"^x + 5"e"^-x) = "a" + "b" ((4"e"^x - 5"e"^-x))/(4"e"^x + 5"e"^-x)`,

Giving 3ex – 5e –x = a(4ex + 5e–x) + b(4ex – 5e–x).

Comparing coefficients on both sides,

We get 3 = 4a + 4b and –5 = 5a – 5b.

This verifies a = `(-1)/8`, b = `7/8`.

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Solved Examples [पृष्ठ १५९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 22 | पृष्ठ १५९

संबंधित प्रश्न

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

\[\int\limits_0^\infty e^{- x} dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^\infty e^{- x} dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals


\[\int\limits_0^1 \tan^{- 1} x dx\]


\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]


\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]


\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Choose the correct alternative:

Γ(n) is


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×