_{0}

^{2 }(x

^{2}dx)/(square root of x

^{3}+1)

_{0}

^{1 }x(square root of x+1)

_{-1}

^{1 }(dx)/(4x

^{2}+1)

_{0}

^{1 }(xdx)/(square root of 1-x

^{4})

_{ }

_{0}

^{ln2 }(e

^{x}dx)/(1+e

^{2x})

1. Use a u-substitution to evaluate the definite integral. (I don't know how to enter the integral sign, so I'm using "S" for the integral sign instead)

(a) S_{0}^{2 }(x^{2}dx)/(square root of x^{3}+1)

(b) S_{0}^{1 }x(square root of x+1)

(c) S_{-1}^{1 }(dx)/(4x^{2}+1)

(d) S_{0}^{1 }(xdx)/(square root of 1-x^{4})_{
}

(e) S_{0}^{ln2 }(e^{x}dx)/(1+e^{2x})

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I will give you the substitutions; I'm sure you can figure it out from there:

(a) u=x^{3}

(b) u=x+1

(c) u=2x

(d) u=x^{2}

(e) u=e^{x}.

Hope this helps! :)

Meg,

As tutors, we are not allowed to solve your problems for you. However, using u-substitution can be simpler if you ensure that you choose the correct value for u. In most cases, you can find u by:

- Using the values within a square root
- Using complex exponents (such as 5
^{3x}) - Or identifying which value within your equation has a derivative which is also included in the equation.

For example, if you want to find the integral of 2e^{2x} dx, you could say that u=2x (in this case the exponent).

You must then solve for the derivative of u, du: du=2dx.

Notice that your original function already has a dx in it, so solve for dx: dx=du/2.

We can now substitute so that the integral of the original function equals: the integral of 2/2e^{u} du.

This works out conveniently, because 2/2=1 and the derivative of e^{u} equals e^{u}.

Therefore, we have e^{u}+C.

We plug in the value of u to get our final answer: e^{2x}+C.

If, as in your problems, you are asked to find the integral between two points, you plug in each point and subtract the upper limit from the lower limit. If finding the integral of our example function between 2 and 0:

e^{2(2)}-e^{2(0)}= e^{4}-e^{0}=54.598-1=53.598

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