Two Curve Area Problem

[Jason76]

POST EDITED

For what value of b is the area of R = area of S? b is a positive constant. Answer: 1.404 How did it get there?

Area R - Below the Curve and Above x Axis

\(\displaystyle y = x\)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} x dx = \dfrac{(b)^{2}}{2} - \dfrac{(0)^{2}}{2} = \dfrac{b^{2}}{2}\)

Area S - Below the Curve and Above x Axis

\(\displaystyle y = \cos(x) \)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} \cos(x) dx = \dfrac{\sin^{2}(b)}{2} - \dfrac{\sin^{2}(0)}{2} = \dfrac{\sin^{2}(b)}{2}\)

 

[lookagain]

For what value of b is the area of R = area of S? b is a positive constant. Answer: 1.404 How did it get there?

Area R - Below the Curve and Above x Axis

\(\displaystyle y = x\)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} x = \dfrac{(b)^{2}}{2} - \dfrac{(0)^{2}}{2} = \dfrac{b^{2}}{2} \ \ \ \ \)You're missing the "dx." *

Area S - Below the Curve and Above x Axis

\(\displaystyle y = \cos(x)\)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} \cos(x) = \ \ \ \ \ \)See * above.

\(\displaystyle \dfrac{\cos^{2}(b)}{2} - \dfrac{\cos^{2}(0)}{2} = \ \ \ \ \)No, that is not the antiderivative for cos(x). You should look that up.


\(\displaystyle \dfrac{\cos^{2}(b)}{2} - \dfrac{1}{2} \ \ \ \ \ \)Then, this line doesn't follow. **

** And so on does not count.

 

[Jason76]

** And so on does not count.

Original Post Edited.

 

[DrPhil]

Original Post Edited.

You can remember the "dx" if you recognize that integration is adding up incremental areas, that is, the integral is a sum of rectangles with height y(x) and width dx.

You still don't have the correct antiderivative for \(\displaystyle \cos x\). What function when differentiated gives \(\displaystyle \cos x\)?

 

[HallsofIvy]

POST EDITED

For what value of b is the area of R = area of S? b is a positive constant. Answer: 1.404 How did it get there?

Area R - Below the Curve and Above x Axis

\(\displaystyle y = x\)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} x dx = \dfrac{(b)^{2}}{2} - \dfrac{(0)^{2}}{2} = \dfrac{b^{2}}{2}\)

Area S - Below the Curve and Above x Axis

\(\displaystyle y = \cos(x) \)

\(\displaystyle x = b, x = 0\)

\(\displaystyle \int^{b}_{0} \cos(x) dx = \dfrac{\sin^{2}(b)}{2} - \dfrac{\sin^{2}(0)}{2} = \dfrac{\sin^{2}(b)}{2}\)

Again, this is wrong. The integral of cos(x) is NOT \(\displaystyle sin^2(x)/2\). Look it up in a calculus book!

 

[Jason76]

Again, this is wrong. The integral of cos(x) is NOT \(\displaystyle sin^2(x)/2\). Look it up in a calculus book!

The integral of \(\displaystyle cos\) is \(\displaystyle sin\), but when evaluating the definite integral you use the "integral power rule". So when using the "integral power rule", how should it come out?

 

[DrPhil]

The integral of \(\displaystyle cos\) is \(\displaystyle sin\), but when evaluating the definite integral you use the "integral power rule". So when using the "integral power rule", how should it come out?

Nonsense. The only difference between an indefinite integral and one with stated limits is that instead of an unknown "constant of integration" you evaluate at the limits and subtract.

You don't have an integer power, you have a cosine. If you expand the cosine as a Taylor series of integer powers, you get

\(\displaystyle \displaystyle \int \cos x \ dx = \int \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdot \cdot \cdot\right)\ dx\)

...................\(\displaystyle =\).....\(\displaystyle \displaystyle \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdot \cdot \cdot\right)+\ C\)

\(\displaystyle \displaystyle \int_0^b \cos x \ dx =\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdot \cdot \cdot\right]_0^b\)

...................\(\displaystyle =\).....\(\displaystyle \displaystyle b - \frac{b^3}{3!} + \frac{b^5}{5!} + \frac{b^7}{7!} + \cdot \cdot \cdot\)

which just happens to be the expansion of \(\displaystyle \sin x\) or \(\displaystyle \sin b\). So you can integrate using integer powers, but most of us just memorize the fact that the integral of a cosine is a sine.

 

[Jason76]

Nonsense. The only difference between an indefinite integral and one with stated limits is that instead of an unknown "constant of integration" you evaluate at the limits and subtract.

You don't have an integer power, you have a cosine. If you expand the cosine as a Taylor series of integer powers, you get

\(\displaystyle \displaystyle \int \cos x \ dx = \int \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdot \cdot \cdot\right)\ dx\)

...................\(\displaystyle =\).....\(\displaystyle \displaystyle \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdot \cdot \cdot\right)+\ C\)

\(\displaystyle \displaystyle \int_0^b \cos x \ dx =\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdot \cdot \cdot\right]_0^b\)

...................\(\displaystyle =\).....\(\displaystyle \displaystyle b - \frac{b^3}{3!} + \frac{b^5}{5!} + \frac{b^7}{7!} + \cdot \cdot \cdot\)

which just happens to be the expansion of \(\displaystyle \sin x\) or \(\displaystyle \sin b\). So you can integrate using integer powers, but most of us just memorize the fact that the integral of a cosine is a sine.

In the case of say:

\(\displaystyle \int^{1}_{2} x dx\)

\(\displaystyle \dfrac{1^{2}}{2} - \dfrac{2^{2}}{2} = \dfrac{1}{2} - \dfrac{4}{2} = -\dfrac{3}{2}\)

:!: But when taking the definite integral of a trig function you do NOT use the integral power rule?

 

[DrPhil]

In the case of say:

\(\displaystyle \int^{1}_{2} x dx\)

\(\displaystyle \dfrac{1^{2}}{2} - \dfrac{2^{2}}{2} = \dfrac{1}{2} - \dfrac{4}{2} = -\dfrac{3}{2}\)

:!: But when taking the definite integral of a trig function you do NOT use the integral power rule?

You ONLY use the integer power rule when you have an integer power.

\(\displaystyle \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1} + C\)

Are you getting confused by trying to find an inverse of the Chain Rule of differentiation?

\(\displaystyle \displaystyle \frac{d}{dx} \cos^nx = n\ \cos^{n-1} x\ \frac{d}{dx}\cos x = n \cos^{n-1}x\ \sin x\)

If you want to integrate a power of cosx, the integrand has to include a factor of sinx.

\(\displaystyle \displaystyle \int \cos x\ \sin x\ dx = \frac{1}{2} \cos^2x + C\)

Are you ready to solve for b yet?