English

Find d∫10-4x+4x2 dx

Advertisements
Advertisements

Question

Find `int sqrt(10 - 4x + 4x^2)  "d"x`

Sum
Advertisements

Solution

We have I = `int sqrt(10 - 4x + 4x^2)  "d"x`

= `int sqrt((2x - 1)^2 + (3)^2)  "d"x`

Put t = 2x – 1

Then dt = 2dx.

Therefore, I = `1/2 int sqrt("t"^2 + (3)^2)  "dt"`

= `1/2 "t" sqrt("t"^2 + 9)/2 + 9/4 log|"t" + sqrt("t"^2 + 9)| + "C"`

= `1/4(2x - 1) sqrt((2x - 1)^2 + 9) + 9/4 log|(2x - 1) + sqrt((2x - 1)^2 + 9)| + "C"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Solved Examples [Page 153]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 7 Integrals
Solved Examples | Q 14 | Page 153

RELATED QUESTIONS

\[\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

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

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

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

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

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

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

\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

If f(x) is a continuous function defined on [−aa], then prove that 

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

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

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

Evaluate each of the following integral:

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

 


If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


Evaluate : 

\[\int\limits_2^3 3^x dx .\]

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

\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\] 

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


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


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


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


\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]


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


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


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


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


Find : `∫_a^b logx/x` dx


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

Γ(1) is


Choose the correct alternative:

If n > 0, then Γ(n) is


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


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


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


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×