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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

∫e4x-3dx = ______ + c - Mathematics and Statistics

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प्रश्न

`int"e"^(4x - 3) "d"x` = ______ + c

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उत्तर

`int"e"^(4x - 3) "d"x` = `bbunderline(1/4e^(4x-3)` + c

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पाठ 1.5: Integration - Q.2

संबंधित प्रश्‍न

Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`


If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:

(A) 0

(B) π

(C) π/2

(D) π/4


If u and v are two functions of x then prove that

`intuvdx=uintvdx-int[du/dxintvdx]dx`

Hence evaluate, `int xe^xdx`


Integrate the function in xlog x.


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


Integrate the function in x sec2 x.


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


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


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


Choose the correct options from the given alternatives :

`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =


Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`


Integrate the following w.r.t. x: `(1 + log x)^2/x`


Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx


Evaluate: `int "dx"/(5 - 16"x"^2)`


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


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


`int ("e"^xlog(sin"e"^x))/(tan"e"^x)  "d"x`


Choose the correct alternative:

`int ("d"x)/((x - 8)(x + 7))` =


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


Evaluate `int 1/(4x^2 - 1)  "d"x`


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


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


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


`int logx  dx = x(1+logx)+c`


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


Complete the following activity:

`int_0^2 dx/(4 + x - x^2) `

= `int_0^2 dx/(-x^2 + square + square)`

= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`

= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`

= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`


Evaluate the following.

`intx^3/sqrt(1+x^4)  dx`


Evaluate the following.

`intx^2e^(4x)dx`


Evaluate the following.

`intx^3 e^(x^2)dx`


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