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The correct answer is sqrt((4x-1)^3)/6, but I always get

2sqrt((4x-1)^3)/3... Can someone explain how to solve that integral pls.

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- Thread starter PrudensOptimus
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- #1

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The correct answer is sqrt((4x-1)^3)/6, but I always get

2sqrt((4x-1)^3)/3... Can someone explain how to solve that integral pls.

- #2

Tom Mattson

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Originally posted by PrudensOptimus

I try to take the integral of sqrt(4x-1) with respect to x...

OK

The correct answer is sqrt((4x-1)^3)/6,

No, it isn't.

Edit: My mistake; yes it is.

but I always get 2sqrt((4x-1)^3)/3...

That's not right either.

Can someone explain how to solve that integral pls.

Yes, you do a u-substitution. Let u=4x-1, so du=4dx. You then have:

(1/4)∫u

which is elementary.

Last edited:

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- #4

HallsofIvy

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The problem is that "4x-1" inside the square root. To get rid of that you make the substitution mentioned earlier: u= 4x-1 so that

√(4x-1)= &radi;(u)= u

du/dx= 4 so du= 4 dx or (1/4)du= dx.

∫ &radic(4x-1)dx= (1/4)&int u

The power rule says that an anti-derivative of u

1/(n+1) u

Since the original problem was given in terms of √ rather than a 1/2 power, it might be a good idea to set the answer in those terms: ∫ &radic(4x-1)dx= (2/3)(√(4x-1))

- #5

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Why can't you just int. the bracket like you would int. x^0.5?

Code:

```
∫ (4x-1)^0.5 dx
increase the power and divied by it:
(4x-1)^1.5 + c = 2(√(4x-1))^3 + c
---------- ------------
1.5 3
```

Last edited:

- #6

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Why can't you just int. the bracket like you would int. x^0.5?

Because if you differentiate your result ((2/3)(4x-1)^(3/2) + C) you get:

4(4x - 1)

which is not what you integrated.

- #7

Tom Mattson

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Originally posted by lavalamp

Why can't you just int. the bracket like you would int. x^0.5?

Code:`∫ (4x-1)^0.5 dx increase the power and divied by it: (4x-1)^1.5 + c = 2(√(4x-1))^3 + c ---------- ------------ 1.5 3`

As Sting said, when you differentiate your result, you do not get the original integrand. That is because (4x-1)

f(u)=u

u(x)=4x-1

In general, a composition of functions does not satisfy the same basic integration rule as the simple function f(u).

edit: typo

- #8

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∫ f'(x) * (f(x))^n dx

in this case f(x)=4x-1 and f'(x)=4. The mulitple in this case is 0.25 which you can take out of the integral and put it back in at the end to get the right answer:

Code:

```
1 * (4x-1)^1.5 + c = (√(4x-1))^3 + c
- ---------- ------------
4 1.5 6
```

- #9

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That's how I learned to do it, way back in '87. I think it's much quicker and easier.

Aaron

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