# Math Help - [SOLVED] Verify an inequality/integral value

1. ## [SOLVED] Verify an inequality/integral value

Without doing any calculus of an integral containing a sine, verify that $\int_0^{\frac{\pi}{2}}xsin(x)\leq\frac{\pi^2}{8}$. My idea was by bounding the integral to get something greater but still lesser than $\frac{\pi^2}{8}$. So I thought about bounding $\int_0^{\frac{\pi}{2}}xsin(x)$ this way : $\int_0^{\frac{\pi}{2}}xsin(x)\leq \int_0^{\frac{\pi}{2}}x=\frac{\pi}{2}$, which is unfortunately greater than $\frac{\pi^2}{8}$. I have no other idea of how to procede. Any idea is welcome.

2. Originally Posted by arbolis
Without doing any calculus of an integral containing a sine, verify that $\int_0^{\frac{\pi}{2}}xsin(x)\leq\frac{\pi^2}{8}$. My idea was by bounding the integral to get something greater but still lesser than $\frac{\pi^2}{8}$. So I thought about bounding $\int_0^{\frac{\pi}{2}}xsin(x)$ this way : $\int_0^{\frac{\pi}{2}}xsin(x)\leq \int_0^{\frac{\pi}{2}}x=\frac{\pi}{2}$, which is unfortunately greater than $\frac{\pi^2}{8}$. I have no other idea of how to procede. Any idea is welcome.
Take a look at the graph. The red curve is $x~sin(x)$ and the blue triangle is square. (The triangle is the area enclosed by the line y = x, the line y = 0 and x = pi / 2.)

What is the area of the triangle?

-Dan

3. Originally Posted by topsquark
What is the area of the triangle?
Oh, I know, teacher pick me, no pick me, I know, can I say?

It is, $\frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{8}$.

4. Take a look at the graph. The red curve is and the blue triangle is square. (The triangle is the area enclosed by the line y = x, the line y = 0 and x = pi / 2.)

What is the area of the triangle?
If I'm not wrong, it's $\frac{\pi^2}{8}$! Nice graphic. It verifies graphically the relation we must verify. Now I'm asking if it's possible to verify it algebraically or any other way that respect the condition "Without doing any calculus of an integral containing a sine", or course not a cosine, or something similar. (I'm asking this because I'm more than sure it's what the expect from us to do).

5. Originally Posted by arbolis
If I'm not wrong, it's $\frac{\pi^2}{8}$! Nice graphic. It verifies graphically the relation we must verify. Now I'm asking if it's possible to verify it algebraically or any other way that respect the condition "Without doing any calculus of an integral containing a sine", or course not a cosine, or something similar. (I'm asking this because I'm more than sure it's what the expect from us to do).
The proof is in post #1. The only mistake is that an anti-derivative of $x$ is $\frac{x^2}{2}$ rather than $x$.

6. The proof is in post #1. The only mistake is that an anti-derivative of is rather than .
Ahahahaha!!!!!!!!! You're right, what an error from my part!!! Thanks.