How do I fix my definite integral issue?

In summary, the given integral is solved by first using substitution to transform it into a form that can be solved using integration by parts. The bounds of integration are also transformed to match the new variable. After applying the fundamental theorem, the final answer is (7 pi)/4-7/2.
  • #1
tangibleLime
71
0

Homework Statement



Integrate from 0 to sqrt(pi/2). I can't figure out how to make the LaTeX work correctly for it, so it just looks like an indefinite integral.

[tex]\int 7\theta^3cos(\theta^2) d\theta[/tex]​


Homework Equations



[tex]\int udv = uv - \int vdu[/tex]​


The Attempt at a Solution



First I took the 7 out of the integral and pulled a theta out of theta cubed in order to match it with the theta squared.

[tex]7 \int \theta^2cos(\theta^2)\theta d\theta[/tex]​

Then I substituted [tex]w = \theta^2, dw = 2\theta d\theta, \frac{dw}{2}=\theta d\theta[/tex].

To make the bounds of integration to work with this substitution...

[tex]w = 0^2 = 0[/tex]
[tex]w = \sqrt{pi/2} = \frac{pi}{2}[/tex]​


After pulling out the 2 that comes along with the dw, this is what my integral is looking like.

[tex]7\frac{1}{2}\int w*cos(w) dw[/tex]​


Now, using integration by parts...

[tex]u = w[/tex]

[tex]du = dw[/tex]

[tex]v = sinw[/tex]

[tex]dv = cosw dw[/tex]

[tex]uv - \int vdu[/tex]
[tex]7\frac{1}{2}(w*sinw - \int sinw dw)[/tex]
[tex]7\frac{1}{2}(w*sinw + cosw)[/tex]​


Substituting back the original value of w...

[tex]7\frac{1}{2}(\theta^2*sin\theta^2 + cos\theta^2)[/tex]​

Now using the Fundamental Theorem...

[tex]7\frac{1}{2}(\frac{pi}{2}^2*sin\frac{pi}{2}^2 + cos\frac{pi}{2}^2) - 7\frac{1}{2}(0^2*sin0^2 + cos0^2)[/tex]​

This gives some ridiculous answer:

[tex]3.5 ((pi/2)^2 sin((pi/2)^2)+cos((pi/2)^2))-3.5 (0^2 sin(0^2)+cos(0^2)) = -3.5 cosh(0)+3.5 (cosh(-i (pi/2)^2)-cos(pi/2+(pi/2)^2) (pi/2)^2)[/tex]​

...which is obviously incorrect. Where did I go wrong?

Thanks.
 
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  • #2
tangibleLime said:
To make the bounds of integration to work with this substitution...

[tex]w = 0^2 = 0[/tex]
[tex]w = \sqrt{pi/2} = \frac{pi}{2}[/tex]​


[tex]7\frac{1}{2}(w*sinw + cosw)[/tex]

Substituting back the original value of w...

Since you've already transformed the limits of integration, there's no reason to substitute back to the original variable. If you do, you need to use the original limits, not the transformed ones.
 
  • #3
Ahhh, thanks. Worked out well.

[tex]((7/2)((pi/2)sin(pi/2)+cos(pi/2)))-((7/2)(0*sin0+cos0))[/tex]

= (7 pi)/4-7/2
 

FAQ: How do I fix my definite integral issue?

What is a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve between two points on a graph. It is represented by the symbol ∫ and is a fundamental tool in calculus.

How is a definite integral different from an indefinite integral?

A definite integral has specific upper and lower limits, while an indefinite integral does not. This means that the result of a definite integral is a single numerical value, while the result of an indefinite integral is an equation with a constant term.

What is the purpose of calculating a definite integral?

The main purpose of calculating a definite integral is to find the area under a curve. This can be useful in a variety of real-world applications, such as calculating the volume of a shape or finding the average value of a function.

What are some common techniques for solving definite integrals?

Some common techniques for solving definite integrals include substitution, integration by parts, and partial fraction decomposition. It is important to have a good understanding of these techniques and when to use them in order to solve more complex definite integrals.

Are there any limitations to using definite integrals?

One limitation of definite integrals is that they can only be used to calculate the area under continuous curves. They cannot be used for functions with discontinuities or infinite intervals. In addition, some integrals may be impossible to solve using traditional methods, requiring more advanced techniques or numerical approximations.

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