Integrating Polynomials: Practice Problems

In summary, the conversation discusses the integration of two equations: 2. /int 1_0 (5u^7+pi^2) dx and 3. /int 4_0 (x^(1/2))(x+1). The answer for 2. is (5/8)+pi^2 and for 3. is 272/15. The conversation also addresses the issue of integrating a constant, pi^2, and the importance of considering the variable of integration, x. The method used to integrate 3. is questioned and eventually corrected.
  • #1
cd246
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Homework Statement


2. /int 1_0 (5u^7+pi^2) dx the answer is (5/8)+pi^2
3./int 4_0 (x^(1/2))(x+1) the answer is 272/15.

Homework Equations





The Attempt at a Solution


For 2. I already have the 5/8, my question do I integrate the pi^2? I tried integrating that with no success.

For 3. (x^(3/2)/3/2)((x^2/2)+x)
then (2x^(3/2)/3)((1/2)x^2+x)
plug 4: (16/3)(8+4=12)=192/3=64
 
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  • #2
for 2: yes you integrate .Pi/2 is a constant.How do you integrate a constant.
for 3: your method is not acceptable.

If you take derivative of (2x^(3/2)/3)((1/2)x^2+x) you don't obtain
(x^(1/2))(x+1) But you should have done so.

So Try to expand (x^(1/2))(x+1) ( get rid of the paranthesis ) Then apply the rule you know about this type of integration.

note:
you can't integrate two parts seperately. Remember for example how do you take derivative of product of functions. it is NOT just the product of derivatives of functions themselves.
 
  • #3
How did you try?

/int 1_0 (5u^7+pi^2)dx ; where is the x? The variable of integration should be x..
Anyway, I assume that was a type-error.

pi^2 is just a constant...

for ./int 4_0 (x^(1/2))(x+1)

I assume you mean:

/int 4_0 (x^(1/2))(x+1)dx

Have you tried to multiply (x^(1/2)) into (x+1) ?
 
  • #4
I found what i did wrong now. thanx
 

What is definite integration 2?

Definite integration 2 is a method used in calculus to find the exact area under a curve between two specific points. It is also known as Riemann integration.

What is definite integration 3?

Definite integration 3 is an extension of definite integration 2, where the interval of integration is divided into smaller subintervals to get a more accurate approximation of the area under a curve. This method is also known as the Trapezoidal Rule or Simpson's Rule.

What is the difference between definite integration 2 and 3?

The main difference between these two methods is that definite integration 3 uses smaller subintervals to calculate the area under a curve, resulting in a more accurate approximation compared to definite integration 2. This makes it a more advanced and precise method for finding the area under a curve.

What are some real-life applications of definite integration 2 and 3?

Definite integration 2 and 3 have various real-life applications in fields such as physics, engineering, and economics. For example, calculating the work done by a variable force, finding the center of mass of a three-dimensional object, and estimating the revenue of a company are all problems that can be solved using definite integration 2 and 3.

How can I improve my skills in definite integration 2 and 3?

To improve your skills in definite integration 2 and 3, it is important to have a strong understanding of the fundamental concepts of calculus. Practice solving problems and familiarize yourself with different techniques and formulas used in these methods. You can also seek help from a tutor or join a study group to further enhance your understanding.

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