Question

Area Function and Integrals

Answer 1

Integrals are a breeze for me to solve, but graphing them is another thing. Why is it that when I graph part a) on Wolfram Alpha, it only goes as far as the [-4, 4] ? So, I would need to sketch the graph of f(x) from [-6, 6], not the integral, is that correct?

Also, does part b) refer to the integral of f(x) and sketching that graph from 0 to x? If so, wouldn't I get an individual value, because it is a definite integral?

And to summarize: the graph for part a) is the function from -6 to 6, and the graph for part b) is the function after it is integrated from 0 to x, is that right?

Answer 2

the derivatiive of the integral is the integrand
the derivative of your integeral is
\[\cos(\frac{\pi x^2}{2})\]

Answer 3

I think the domain that WA uses for plotting is just whichever gives the viewer a good idea of what the function looks like. If you want to manually input your own domain, type the following:
\[\text{Plot[$f(x)$,}\{x,a,b\}]\]
where \(f(x)\) is your function and \([a,b]\) is the interval.

Answer 4

http://www.wolframalpha.com/input/?i=cos%28pi+x^2%2F2%29+domain+-6..6

Answer 5

Thanks for your answers...

But thinking that the graph of \[\int\limits_{-6}^{6} \cos(\frac{ \pi t^2 }{ 2 })\]

is different from the graph of \[\cos(\frac{ \pi t^2 }{ 2 })\]

Plus, I had other questions above as well.

Answer 6

Correction: *But I'm thinking

Answer 7

\[\int\limits_{-6}^{6} \cos(\frac{ \pi t^2 }{ 2 })dx\] is a number

Answer 8

Or is that just the range?

Answer 9

For part (a), the function \(f(x)\) is a function that tells you the area under the curve \(\cos\dfrac{\pi t^2}{2}\) from a fixed point (0) to some \(x\) as \(x\) changes.

Over the desired interval, \([-6,6]\), you have endpoints
\[\int_0^{-6}\cos\frac{\pi t^2}{2}~dt=-\int_{-6}^0\cos\frac{\pi t^2}{2}~dt\]
and
\[\int_0^{6}\cos\frac{\pi t^2}{2}~dt\]
which are both numbers, both areas. Now, seeing as there's no elementary method for computing such an integral (unless you're comfortable with a power series approximation), I don't understand how the question would expect you to graph this function.

Interestingly, according to WA, there's a function defined by this same exact integral:

http://mathworld.wolfram.com/FresnelIntegrals.html

(\(C(x)\) is the one you have).

Answer 10

Part (b) is an application of the first fundamental theorem of calculus, which says
\[\frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\cdot g'(x)\]
where \(c\) is some constant in the domain of \(f\). As satellite mentioned above, the derivative would be
\[f'(x)=\cos\frac{\pi x^2}{2}\]
which should be easier to graph than its antiderivative \(f(x)\).

Answer 11

@SithsAndGiggles: I'm understanding what you're saying....

Answer 12

Consider an example,
\[p(x)=\int_0^x dt\]
And say we're concerned with what this function looks like over \([-2,2]\).

At \(x=-2\):
\[p(-2)=\int_0^{-2}dt=-\int_{-2}^0dt=-(0-(-2))=-2\]
At \(x=2\):
\[p(2)=\int_0^{2}dt=2-0=2\]
Go ahead and see for yourself what happens if you check values between -2 and 2. Here's a plot of \(p(x)\):

Answer 13

The function \(p(x)\) gives the area under \(Q(x)=1\).

For \(p(-2)\), the "area" is negative because we're travelling in the negative direction (0 to -2).