integral help?

Question

can anyone tell me what i need to do to evaluate the integral of (4x-3)^9dx tell me exactly what ur doing please

why am I getting 1/40(4x-3)^10+c as answer in the back of the book

why am I getting 1/40(4x-3)^10+c as answer in the back of the book

Answer 1

increase the power by one and divide the whole thing by the current power
so its (4x-3)^10/9

its backward of derivative formula
n(ax)^(n-1) for D
and for integral

1/n(ax)^n+1

Answer 2

For Int (4x-3)^9 dx, call z = 4x-3 and integrate in terms of z with a conversion factor:

Int (4x-3)^9 dx = Int z^9 dz *dx/dz

but dz/dx is simple, equals 4. So dx/dz = 1/4

Int (4x-3)^9 dx = Int 1/4 z^9 dz

= 1/4 * { 1/10 * z^10 + C} where C is any constant.

= 1/40 z^10 + 1/4*C.

= 1/40 (4x-3)^10 + 1/4C.

Prove it by differentiating again, call the above f(x)

df(x)/dx = df(x)/dz * dz/dx = 1/40 {10*z^9} * 4 = z^9 = (4x-3)^9.

Answer 3

Integral [(4x-3)^9] dx

= Integral [Z^9] dx

where Z = 4x - 3

Since dZ = 4dx, dx = dZ/4, so the integral is also, by change of variable of integration,

Integral [(Z^9)/4] dZ

= (1/4) Integral [Z^9] dZ
= (1/4) (Z^10)/10
= (1/40) (Z^10)
= (1/40) (4x - 3)^10

and they always add the constant for the indefinite integral.

Answer 4

first u=4x-3 so the derivative du=4dx...since the coefficient of the problem is 1, you have to make 4 into 1...you do this by deividing 4 by 1 which is 1/4...that makes the other side 1/4du....so its 1/4du=1....subsitute 1/4du in for the coefficient 1 to make your equation 1/4*integral(u^9)du...now just integrate u^9, and replace u with u=4x-3....and you have your answer....don't forget +C if you aren't finding an exact answer!

The pervious people's answers are wrong because they forgot about the (4x-3) which needs integrating as well.

Answer 5

Integral of (4x-3)^9 dx

Use the power rule, which states (1/n+1)(u)^n+1
u=4x-3
n=9
1/10(4x-3)^10

Then I think you should add a constant.

1/10(4x-3)^10+C

Not garanteed to be correct!

Answer 6

Let u = 4x - 3
du / dx = 4
dx = du / 4
I = (1/4).∫ (u)^9.du
I = (1/40).u^10 + C
I = (1/40).(4x - 3)^10 + C

Answer 7

integ (4x -3)^9 dx
= (4x -3)^10 (4 )/10 + c
= (2 )(4x -3)^10 / 5 + c
= 0.4 (4x -3)^10 + c