# Evaluate using u-substitution?

Evaluate using u-substitution when u= x+1

Go slow so I can figure out what you're doing please, that's the only way I'm going to learn.

(∫ from a to b) x^2 (√x+1) dx

when a = 0 and b= 1

both the x and the 1 are under the square root

### 3 Answers

- MadhukarLv 71 decade agoFavorite Answer
I assume you mean x^2 √(x + 1) and not x^2 √(x + 1). Both are different.

Let x + 1 = u

=> dx = du and

x = u - 1

Let us first perform indefinite integration by u-substitution, then change the answer in terms of x before plugging in limits of x. Thus,

∫ x^2 √(x + 1) dx

= ∫ (u - 1)^2 * √u du

= ∫ (u^2 - 2u + 1) * u^(1/2) du

= ∫ [u^(5/2) - 2u^(3/2) + u^(1/2)] du

= (2/7) u^(7/2) - 2 * (2/5) u^(5/2) + (2/3) u^(3/2) + c

= (2/105) u^(3/2) (15u^2 + 21u + 35) + c

Plugging back u = x+1,

= (2/105) (x+1)^(3/2) [15(x+1)^2 - 42(x+1) + 35) + c

= (2/105) (x+1)^(3/2) (15x^2 - 12x + 8) + c

Plugging limits 0 to 1,

= (2/105) [2^(3/2) (15 - 12 + 8) - 8]

= (4/105) (11√2 - 8).

Edit:

Revised, simplified and corrected.

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- JoshLv 61 decade ago
Unlike differential calculus, there is no chain rule in integral calculus. So when you have two functions multiplied, there is no set rule for how to integrate them. Infact there is very little in the way of interesting functions that can be integrated, most of the interesting problems can not be integrated in any simple fashion.

Keep in mind that basically any time you see a variable in a piece of math, it can be considered a function.

The idea behind u substition is to simplify the the expression into somthing we do know how to integrate. It does this by using what is called a "change in variable." Bacisally, all you are doing is taking the problematical section out of the expression, and calling it something else.

This takes a bit of practice and artistry. The practice is important because you need to have some idea of what integral calculus answers look like, so that you can chose a part of the expression to replace. Generaly this will be the stuff inside a root, inside an exponent, or the stuff on the bottom of a fraction, though the stuff ontop is also fairly common.

Lets take a quick step back, so we can see how we got where we are now.

take the generic function

y = x^2

if take the derivative, we get

dy/dx = 2x

now you should know that dy/dx is not really a fraction, but for this excercise we can treat it as such

dy = 2xdx

this implies explicitly that "the derivative of y is two times x, with respect to x."

now if we intergrate both sides, we get

y = x^2 + C

so that dx, is an inherent part of the expression. The over-arching implication here is that any function, is infact the derivative of another function.

This implies that we can use that dx algebraicaly at times.

so we need to get rid of part of the math we dont know how to integrate, and we need to get rid of the dx as well.

∫x^2rt(x+1)dx

let u = 1+x

take the derivative of both sides

du = 1dx

so we can replace x+1 with u, and dx with du

∫x^2du

now since u = x+1

u = x+1

u-1 = x

sub that in

∫(u-1)^2du

∫(u^2 - 2u +1)du

this we know how to integrate

1/3u^3 - u^2 + u

1/3(x+1) -(x+1)^2 + (x+1)| from a to b

[1/3 - 1 + 1] - [2/3 - 4 + 2]

[1/3] - [-4/3] = 2

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- 1 decade ago
When u=x+1, we have

1. x=0 corresponds to u=1, x=1 corresponds to u=2,

2. x=u-1, so x^2=(u-1)^2=u^2-2u+1

3. dx=du.

Then the problem becomes: integrate (u^2-2u+1)*sqrt(u) from u=1 to u=2.

i.e. integrate (u^5/2-2u^3/2+u^1/2) term by term, evaluate at u=2, u=1, then subtract.

The answer should be nearly 0.4402

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