Calc 2: Integral of 1/sqrt(4-x^2)

In summary: Exactly that situation. Fortunately my college's Calc II class has the syllabus posted online and it looks like the only discrepancies are trig integrals and substitution and approximate integration. They both cover Maclaurin series, for me. Good luck in your class and thanks again for your help!
  • #1
Jordan P
3
0

Homework Statement


∫ from 0 to 1 of 1/sqrt(4-x^2) dx

I know from Wolfram Alpha that the answer should be π/6, and that the indefinite integral would be arcsin(x/2).
I see the connection between this and the antiderivative of arcsin, I'm just not sure how to handle the 4 rather than the 1--which technique to use.

Homework Equations


∫ 1/sqrt(1-x^2)dx is arcsin x

The Attempt at a Solution


Tried a few ways.

1st attempt was to rewrite as ∫ (4-x^2)^(-1/2)dx then reverse the power and chain rules to get sqrt(4-x^2)/2x from 0 to 1=sqrt(3)/2. Knew this was incorrect but trying to get my thoughts flowing.

2nd attempt was with integration by parts, with u=1/sqrt(4-x^2) and dv=dx...etc. This one ended up pretty ugly though, as vdu was worse than the original equation.

3rd attempt was with u-substitution; this one looked like it was going to work then took a bad turn. Let u=4-x^2, produces ∫ from 4 to 3 1/sqrt(u) --> -sqrt(u) from 4 to 3, then plugging back the 4-x^2 ended up undefined.

I was thinking of partial fractions but the sqrt throws me off. I'm studying for a placement exam and this is a question from a past exam in the class.

Thanks to everyone very much for your help!
 
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  • #2
Write the integrand
[tex]\frac{1}{\sqrt{4-x^2}}=\frac{1}{2}\frac{1}{\sqrt{1-(x/2)^2}}[/tex]
and use t=x/2 as integration variable.

ehild
 
  • #3
∫ from 0 to 1 of 1/sqrt(4-x^2) dx
If you have no idea how to evaluate this just use a trigonometric function substitution, if you haven't memorized what this integral is, have you learned about how to make these substitutions yet? It's something you should of learned or well in most calculus II courses.

This or the option above both work.
 
  • #4
ehild's is a good answer. Another way of approaching it is to use the substitution, as GreenPrint suggests, x= 2sin(u). That way [itex]4- x^2= 4- 4sin^2(u)= 4(1- sin^2(u)= 4 cos(u)[/itex].
 
  • #5
Awesome, thanks guys. Looking at the course's syllabus it looks like they were going for trig substitution here, a topic we didn't cover in my BC class in high school. However, both solutions make sense and now I've learned something new! Thank you again.
 
  • #6
Jordan P said:
Awesome, thanks guys. Looking at the course's syllabus it looks like they were going for trig substitution here, a topic we didn't cover in my BC class in high school. However, both solutions make sense and now I've learned something new! Thank you again.

This is not a good thing. I take it your taking Multivariable Calculus this semester than sense you took the college equivalent of Calculus II in high school? I guess the AP program doesn't directly reflect courses taught in universities. I also noticed that Maclaurin series is taught in AP BC Calculus but not in my college Calculus II course... interesting... the two seem to differ slightly...
 
  • #7
GreenPrint said:
This is not a good thing. I take it your taking Multivariable Calculus this semester than sense you took the college equivalent of Calculus II in high school? I guess the AP program doesn't directly reflect courses taught in universities. I also noticed that Maclaurin series is taught in AP BC Calculus but not in my college Calculus II course... interesting... the two seem to differ slightly...

Exactly that situation. Fortunately my college's Calc II class has the syllabus posted online and it looks like the only discrepancies are trig integrals and substitution and approximate integration. They both cover Maclaurin series, for me. Good luck in your class and thanks again for your help!
 

Related to Calc 2: Integral of 1/sqrt(4-x^2)

What is the mathematical definition of the integral of 1/sqrt(4-x^2)?

The integral of 1/sqrt(4-x^2) is the anti-derivative of the given function, which is represented by ∫(1/√(4-x^2))dx. It is a mathematical concept used to find the area under a curve or the accumulation of a quantity over a given interval.

What is the domain and range of the function 1/sqrt(4-x^2)?

The domain of the given function is all real numbers except for values of x that make the denominator 0. Therefore, the domain is (-2, 2) and the range is [0, ∞).

How do you solve the integral of 1/sqrt(4-x^2)?

To solve the integral of 1/sqrt(4-x^2), you can use the substitution method. Let u = 4-x^2, then du = -2xdx. Substitute these values into the integral to get ∫(1/√u)(-1/2)du. This can be simplified to -1/2∫(1/√u)du, which is equal to -√u + C. Finally, substitute back u = 4-x^2 to get the final solution of -√(4-x^2) + C.

What is the geometric interpretation of the integral of 1/sqrt(4-x^2)?

The integral of 1/sqrt(4-x^2) represents the area under the curve of the given function on a graph. This can be visualized as the area of a semicircle with a radius of 2, centered at the origin on the x-y plane. This is because the function 1/sqrt(4-x^2) is equivalent to the equation of a semicircle in polar coordinates.

What are the real-world applications of the integral of 1/sqrt(4-x^2)?

The integral of 1/sqrt(4-x^2) has many applications in physics, particularly in calculating the work done by a force that varies as 1/sqrt(4-x^2). It is also used in engineering, such as in designing curved structures like arches and bridges. Additionally, it has applications in calculating the area under certain curves in economics and finance.

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