1. ## definite integrals

Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals

Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.

2. ## Re: definite integrals

Originally Posted by sonalig
Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals .

That is simply the area enclosed by the unit circle in the first quadrant.
How much is that?

3. ## Re: definite integrals

Let's think of $x$ and $1$ in the given question as two sides of a triangle like in this image below:

$\cos{(\theta)} = \frac{x}{1} = x$

$\sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$

$\text{And } dx = -\sin{(\theta)}\,\,d\theta$

Now for lower limit $x = 0$:
$x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}$

And for upper limit $x = 1$:

$x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$

Back to the main question which can be written as:

\begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}

4. ## Re: definite integrals

Originally Posted by x3bnm
Let's think of $x$ and $1$ in the given question as two sides of a triangle like in this image below:

$\cos{(\theta)} = \frac{x}{1} = x$
$\sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$
$\text{And } dx = -\sin{(\theta)}\,\,d\theta$
Now for lower limit $x = 0$:
$x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}$
And for upper limit $x = 1$:
$x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$
Back to the main question which can be written as:
\begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}
@x3bnm
Why would anyone changed such a transparently simple question into such a complicated mess?

5. ## Re: definite integrals

I guess I like complicated things . Can you tell me how you would solve it if you don't mind telling me?

6. ## Re: definite integrals

Originally Posted by sonalig
Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals

Can somebody please help me with this one, I am not understanding how to approach it. I know it forms a fourth of a circle because of the bounds but that is all I am understanding.
Surely you know that the area of a circle is calculated by $\displaystyle A = \pi \, r^2$. Here you have a quarter of the circle of radius 1, so what is the area?

7. ## Re: definite integrals

IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

8. ## Re: definite integrals

Originally Posted by agentmulder
IMHO x3bnm has shown very nicely how to work integrals using trigonometric substitutions. The OP asked for sums and differences of areas which x3bnm provided using simple geometric formulas of right triangles.

Surely an integral with trigonometric substitutions is much more difficult than $\displaystyle \frac{\pi \, r^2}{4}$. It's 100% correct and beautifully executed (and LaTeXed), but the question asked to use simple geometric formulas. Surely there are few more simple than the formula for the area of a circle...

9. ## Re: definite integrals

We know that the function of the given question is:

$y = \sqrt{1-x^2}........\text{(1)}$

By polar coordinate we know that $x = r\cos(\theta)$ and $y = r\sin(\theta)$
If we plugin these values into equation (1) we get:

\begin{align*}r\sin{(\theta)} =& \sqrt{1-r^{2}\cos^2{(\theta)}} \\ r^2\sin^2{(\theta)} =& 1 - r^{2}\cos^2{(\theta)}\\ r^2(\sin^2{(\theta)} + \cos^2{(\theta)}) =& 1\\ \therefore\,\, r =& 1\end{align*}

For upper limit $x = 1$
$x = r\cos{(\theta)} = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$

For lower limit $x = 0$

$x = r\cos{(\theta)} = \cos{(\theta)} = 0 \therefore\,\, \theta = \frac{\pi}{2}$

So the area is one fourth of the full quadrant.

$Area = \pi r^2 - \frac{3 \pi r^2}{4} = \frac{\pi}{4}.........\text{[because } r = 1 \text{]}$