English

2 ∫ 0 √ 4 − X 2 D X

Advertisements
Advertisements

Question

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]
Advertisements

Solution

\[\int_0^2 \sqrt{4 - x^2} d x\]
\[ = \int_0^2 \sqrt{2^2 - x^2} d x\]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{1}{2} \times 2^2 \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} \right]_0^2 + 2 \left[ \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = 0 + 2\left( \frac{\pi}{2} - 0 \right)\]
\[ = \pi\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Very Short Answers [Page 115]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Very Short Answers | Q 24 | Page 115

RELATED QUESTIONS

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

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

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

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

 


\[\int\limits_a^b e^x dx\]

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

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


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

If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


\[\int\limits_0^{\pi/4} \tan^4 x dx\]


\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]


\[\int\limits_0^{2\pi} \cos^7 x dx\]


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


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


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


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×