∫(6x-x^2)^1/2dx?

Question

∫(6x-x^2)^1/2dx = ???
How to calculate??

Answer 1

Not sure how to do it but there is a nice utility from Wolfram research that does it for you. Perhaps you can work back from this.

(Sqrt[-((-6 + x)*x)]* (Sqrt[-6 + x]*(-3 + x)*Sqrt[x] - 18*Log[Sqrt[-6 + x] + Sqrt[x]]))/ (2*Sqrt[-6 + x]*Sqrt[x])

Answer 2

enable's see ... 6x + 4(3x - 2) = 6x + 2 6x + 12x - 8 = 6x + 2 6x + 12x - 6x - 8 - 2 = 0 12x - 10 = 0 12x = 10 x = 10/12 x = 5/6 properly, all human beings's answer's the comparable, who you are going to confirm on because of the fact the suitable answer? ask your self why you're asking this.

Answer 3

To solve this problem, you have to complete the square inside the square root (which is what things to the power of one half are). You want to complete the square of

-x^2 + 6x

Using normal steps, you'd first make the coefficient of x^2 positive, by factoring out a -1.

-(x^2 - 6x)

In order to complete the square, you have to take "half squared" of the coefficient of x, and half of -6 is -3, squared is 9. So our magic number is 9 in this case

-(x^2 -6x + 9) + ?

Note the question mark; you have to replace the question mark to offset the value you added in. Since we really added in -9 instead of 9 (due to the brackets), we have to add 9

-(x^2 - 6x + 9) + 9

And we factor accordingly.

-(x - 3)^2 + 9

Which can be written as

9 - (x-3)^2

So our new integral is then
∫sqrt[(9 - (x - 3)^2)] dx

This will ultimately become trig substition, but it isn't obvious yet. First, you have to use U substitution.

Let u = x - 3
du = dx

Subsequent to the substition, we have
∫sqrt[(9 - u^2)] du

And now, this is one of our desired forms for trig substition. We'd normally use the symbol "theta" but with this limited interface I'm going to use y instead.
Let u = 3sin(y)
du = 3cos(y) dy

Giving us

∫sqrt[(9 - (3siny)^2)] 3cosy dy
∫sqrt[9 - 9sin^2(y)] 3cosy dy
∫sqrt[9(1 - sin^2(y))] 3cosy dy

And now, we can pull out the 9 out of the square root and out of the integral to get 3, as well as the other constant.

3*3 ∫sqrt(1 - sin^2(y)) cosy dy

Remember our identity that 1 - sin^2(x) = cos^2(x), so we replace it like so.

9 ∫sqrt(cos^2(y)) cosy dy

Now, we're taking the square root of a square value, so it would just be itself.

9 ∫cos(y)*cos(y) dy
9 ∫cos^2(y) dy

To solve even powers of cosine, you have to use the identity
cos^2(y) = (1 + cos2y)/2. Let's plug that in.

9 ∫(1 + cos2y)/2 dy

And again, let's pull out that pesky constant.

9/2 ∫(1 + cos2y) dy

Now, the integral is easy to solve.

9/2 (y + (1/2)sin2y) + C

Let's convert sin2y with the double angle identity 2sinycosy. The 2 will cancel out the (1/2) next to it, leaving us with just sinycosy

9/2 (y + sinycosy) + C

But remember that we expressed y in terms of u. That is, we let
u = 3siny.

This means that
siny = u/3

At this point, you have to draw a right angle triangle, declaring one of the angles to be y, and noting that by the SOHCAHTOA rules, sin = opp/hyp. Since sin(y) = u/3, make the triangle have "u" as its opposite, and "3" as its hypotenuse. That means the adjacent side is equal to sqrt(9 - u^2) by the Pythagorean Theorem.

Back to this:

9/2 (y + sinycosy) + C

siny = opp/hyp = u/3
To solve for cosy, just use the triangle you created.
cosy = adj/hyp = sqrt(9 - u^2)/3

To solve for y, since u = siny, then y = arcsin(u)

So our answer now becomes

9/2 (arcsin(u) + (u/3)(sqrt(9 - u^2)/3) + C
9/2 ( arcsin(u) + (u/9)(sqrt(9 - u^2) ) + C

BUT our answer is in terms of x. We let u = x - 3

9/2 ( arcsin(x-3) + ((x-3)/9)(sqrt(9 - (x-3)^2) ) + C

And if we really wanted to reduce, we can distribute the 9/2

9/2 arcsin(x-3) + [(x-3)/2](sqrt(9 - (x-3)^2 ) + C

Answer 4

Do it by parts let v = (6x-x²) and du = dx. Then the integral becomes ∫vdu = uv - ∫udv and the result (after simplification) is (-1/6)(x-9)x²


Doug

Answer 5

if ^ means times, the answer is 2/d, because the parentheses make 4x and you divide the resulting fraction by 2x

Answer 6

((6x-x^2)^3/2)/(6-2x)

Answer 7

the answer is cheese sauce ....i think .....oh yes im definetely sure after recalculations....mmm...yas!!!!