मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Integrate the following functions w.r.t. x : (x-3)(7-x)

Advertisements
Advertisements

प्रश्न

Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`

बेरीज
Advertisements

उत्तर

Let I = `int sqrt((x - 3)(7 - x)).dx`

=`intsqrt(-x^2 + 10x - 21).dx`

= `int sqrt(- (x^2 - 10x + 21)).dx`

= `int sqrt(4 -(x^2 - 10x + 25)).dx`

= `int sqrt(2^2 - (x - 5)^2`

= `((x - 5)/2) sqrt(2^2 - (x - 5)^2) + 2^2/(2) sin^-1 ((x - 5)/2) + c`

= `((x - 5)/2) sqrt((x - 3)(7 - x)) + 2sin^-1 ((x - 5)/2) + c`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.3 | Q 2.06 | पृष्ठ १३८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Integrate the function in (sin-1x)2.


Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.


Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.


Evaluate the following:

`int x tan^-1 x . dx`


Evaluate the following:

`int sec^3x.dx`


Integrate the following functions w.r.t.x:

`e^-x cos2x`


Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`


Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`


Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`


If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.


Integrate the following w.r.t.x : cot–1 (1 – x + x2)


Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`


Integrate the following w.r.t.x : log (log x)+(log x)–2 


Solve the following differential equation.

(x2 − yx2 ) dy + (y2 + xy2) dx = 0


Evaluate the following.

`int x^2 e^4x`dx


Evaluate: Find the primitive of `1/(1 + "e"^"x")`


Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`


Evaluate: `int "dx"/("9x"^2 - 25)`


Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx


Evaluate:

∫ (log x)2 dx


`int (sinx)/(1 + sin x)  "d"x`


`int (sin(x - "a"))/(cos (x + "b"))  "d"x`


`int ("d"x)/(x - x^2)` = ______


`int (x^2 + x - 6)/((x - 2)(x - 1))  "d"x` = x + ______ + c


`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.


`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.


`int tan^-1 sqrt(x)  "d"x` is equal to ______.


Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`


Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.


If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.


If `int(x + (cos^-1 3x)^2)/sqrt(1 - 9x^2)dx = 1/α(sqrt(1 - 9x^2) + (cos^-1 3x)^β) + C`, where C is constant of integration , then (α + 3β) is equal to ______.


Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.


Evaluate: 

`int(1+logx)/(x(3+logx)(2+3logx))  dx`


`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`


Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.

Solution: (x2 + y2) dx - 2xy dy = 0

∴ `dy/dx=(x^2+y^2)/(2xy)`                      ...(1)

Puty = vx

∴ `dy/dx=square`

∴ equation (1) becomes

`x(dv)/dx = square`

∴ `square  dv = dx/x`

On integrating, we get

`int(2v)/(1-v^2) dv =intdx/x`

∴ `-log|1-v^2|=log|x|+c_1`

∴ `log|x| + log|1-v^2|=logc       ...["where" - c_1 = log c]`

∴ x(1 - v2) = c

By putting the value of v, the general solution of the D.E. is `square`= cx


`inte^(xloga).e^x dx` is ______


`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`


Evaluate:

`int e^(ax)*cos(bx + c)dx`


`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.


If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv  dx - int(d/dx u)(intv  dx)dx`. Hence evaluate: `intx cos x  dx`


Evaluate the following. 

`int x sqrt(1 + x^2)  dx`  


The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:


Under the LIATE rule, which type of function has priority \(L\)?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×