Integration Questions: Check Answers and Simplify Solutions | Homework Help

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In summary, the conversation discusses integrals and the request for them to be checked and simplified. The first integral involves x to the power of the square root of 2, the second involves a product of x and square root of x, and the third involves x to the power of negative pi. Alternative methods for integration are suggested, including simplifying the integrand and using integration by parts or substitution. The final solution for the first integral is corrected, and the importance of including the dx in integrals is mentioned.
  • #1
roadworx
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Homework Statement



I'm wondering if anyone can check these integrations for me, or suggest alternative answers if they're not quite right, or can be simplified?

1) [tex]\int x^{\sqrt{2}}[/tex]

2) [tex]\int x . \sqrt{x}[/tex]

3) [tex]\int \frac{1}{x^\pi}[/tex]

Homework Equations



--

The Attempt at a Solution



1) [tex]\int x^{1.4} = \frac{x^{2.4}}{2.4} + C [/tex]

2) [tex]\int x . x^{\frac{1}{2}} = \frac{x^2}{2} . \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{x^{\frac{3}{2}}}{3} . x^2 + C [/tex]

3) [tex]\int \frac{1}{x^\pi} = \int x^{-\pi} = \frac{x^{-\pi+1}}{-\pi+1} + C [/tex]
 
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  • #2
The first and the third look okay to me, although in the first one I would leave the [tex]\sqrt{2}[/tex] rather than putting 2.4.

In the second your almost there. When doing this sort of integral it is often easier if you simplify the integrand. By combining [tex]x*\sqrt{x}[/tex] into [tex]\sqrt{x*x^2}[/tex] to get [tex]{x^{3/2}}[/tex] it should be easier to integrate.
 
  • #3
roadworx said:

Homework Statement



I'm wondering if anyone can check these integrations for me, or suggest alternative answers if they're not quite right, or can be simplified?

1) [tex]\int x^{\sqrt{2}}[/tex]

2) [tex]\int x . \sqrt{x}[/tex]

3) [tex]\int \frac{1}{x^\pi}[/tex]

Homework Equations



--

The Attempt at a Solution



1) [tex]\int x^{1.4} = \frac{x^{2.4}}{2.4} + C [/tex]
[itex]\sqrt{2}[/itex] is NOT equal to 1.4! There is no reason to change [itex]x^{\sqrt{2}}[/itex] to [itex]x^{1.4}[/itex].

2) [tex]\int x . x^{\frac{1}{2}} = \frac{x^2}{2} . \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{x^{\frac{3}{2}}}{3} . x^2 + C [/tex]
An unfortunately common mistake: just as you cannot differentiate a product by just differentiating each part, you cannot integrate a product that way either. As Vuldoraq said, [itex]x\sqrt{x}= x(x^{1/2})= x^{3/2}[/itex]. Integrate that.

3) [tex]\int \frac{1}{x^\pi} = \int x^{-\pi} = \frac{x^{-\pi+1}}{-\pi+1} + C [/tex]
Yes, this is correct.
 
  • #4
Thanks for the replies Vuldoraq, HallsofIvy.

So, my answer should be [tex]\int x^{\frac{3}{2}} = \frac{2x^{\frac{3}{2}}}{3} [/tex]
 
  • #5
You forgot to increase the exponent by one before dividing!
 
  • #6
Are you aware of the fact that you're leaving out the dx in the integrals?
 
  • #7
alright..

Second attempt.

[tex] \int x^{\frac{3}{2}} = \frac{2x^{\frac{5}{2}}}{5} [/tex]
 
  • #8
Thats it, much better. :smile:

To repeat what HallsofIvy said: you can't integrate products of the function being integrated in the usual manner. So you have to simplify and when you can't simplify you have to use a different method (like integration by parts or substitution, you'll come across these later on).

May your Math prosper
 

What is integration?

Integration is a mathematical process of finding the area under a curve by breaking it into smaller parts and summing them up.

What is the purpose of integration?

The purpose of integration is to calculate the total value of a function over a given interval, which can be useful in various fields such as physics, engineering, economics, and more.

What are the two types of integration?

The two types of integration are definite integration and indefinite integration. Definite integration involves finding the exact value of the area under the curve, while indefinite integration involves finding the general solution or antiderivative of a function.

What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that differentiation and integration are inverse operations of each other. This means that the derivative of an integral is equal to the original function, and the integral of a derivative is equal to the original function (plus a constant).

What are some real-life applications of integration?

Integration has various real-life applications, such as calculating the distance traveled by an object with a changing velocity, determining the amount of medicine in a person's bloodstream, and finding the area under a demand curve to calculate revenue in economics.

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