मराठी

Write the Coefficient A, B, C of Which the Value of the Integral 3 ∫ − 3 ( a X 2 + B X + C ) D X is Independent.

Advertisements
Advertisements

प्रश्न

Write the coefficient a, b, c of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.
Advertisements

उत्तर

\[\int_{- 3}^3 \left( a x^2 + bx + c \right) d x\]
\[ = \left[ a\frac{x^3}{3} + b\frac{x^2}{2} + cx \right]_{- 3}^3 \]
\[ = 9a + \frac{9}{2}b + 3c + 9a - \frac{9}{2}b + 3c\]
\[ = 18a + 6c\]

Hence, the given integral is independent of b

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 35 | पृष्ठ ११६

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

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

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

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx\]

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

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

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int_{- 2}^2 x e^\left| x \right| dx\]

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


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

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

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

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

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_0^5 \left( x + 1 \right) dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\]  is equal to

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

\[\int\limits_0^{\pi/2} x \sin x\ dx\]  is equal to

Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

\[\int\limits_0^4 x\sqrt{4 - x} dx\]


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]


\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


Evaluate the following:

`int_(-1)^1 "f"(x)  "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x  < 0):}`


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×