Which is larger: I_1 = integral from 0 to 1 x² sin² x dx

Which is larger: $$ I_1 = \int_0^1 x^2 \sin^2 x \, dx \quad \text{or} \quad I_2 = \int_0^1 x \sin^2 x \, dx $$

Method 1: Direct comparison

Consider the difference:

$$I_1 - I_2 = \int_0^1 x^2 \sin^2 x \, dx - \int_0^1 x \sin^2 x \, dx = \int_0^1 (x^2 - x)\sin^2 x \, dx$$

$$= \int_0^1 x(x-1)\sin^2 x \, dx$$

Step 1: Analyze the sign of $x(x-1)$

For $x \in (0, 1)$:

• $x > 0$
• $x - 1 < 0$
• Therefore: $x(x-1) < 0$

Step 2: Analyze $\sin^2 x$

For $x \in (0, 1)$:

$$\sin^2 x > 0$$

Step 3: Determine the sign of the difference

$$x(x-1)\sin^2 x < 0 \quad \text{for all } x \in (0,1)$$

Therefore:

$$I_1 - I_2 = \int_0^1 x(x-1)\sin^2 x \, dx < 0$$

$$\implies I_1 < I_2$$

$$\boxed{\int_0^1 x \sin^2 x \, dx > \int_0^1 x^2 \sin^2 x \, dx}$$

Conclusion: $I_2$ is larger. ✓

Method 2: Direct calculation (verification)

Let's compute both integrals to verify.

Calculating $I_2 = \int_0^1 x\sin^2 x \, dx$

Using $\sin^2 x = \frac{1 - \cos 2x}{2}$:

$$I_2 = \int_0^1 x \cdot \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2}\int_0^1 x \, dx - \frac{1}{2}\int_0^1 x\cos 2x \, dx$$

$$= \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2}\int_0^1 x\cos 2x \, dx = \frac{1}{4} - \frac{1}{2}\int_0^1 x\cos 2x \, dx$$

For $\int x\cos 2x \, dx$, use integration by parts:

Let $u = x$, $dv = \cos 2x \, dx$

Then: $du = dx$, $v = \frac{\sin 2x}{2}$

$$\int x\cos 2x \, dx = \frac{x\sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx = \frac{x\sin 2x}{2} + \frac{\cos 2x}{4}$$

Therefore:

$$\int_0^1 x\cos 2x \, dx = \left[\frac{x\sin 2x}{2} + \frac{\cos 2x}{4}\right]_0^1$$

$$= \left(\frac{\sin 2}{2} + \frac{\cos 2}{4}\right) - \left(0 + \frac{1}{4}\right) = \frac{\sin 2}{2} + \frac{\cos 2 - 1}{4}$$

So:

$$I_2 = \frac{1}{4} - \frac{1}{2}\left(\frac{\sin 2}{2} + \frac{\cos 2 - 1}{4}\right) = \frac{1}{4} - \frac{\sin 2}{4} - \frac{\cos 2 - 1}{8}$$

$$= \frac{1}{4} - \frac{\sin 2}{4} - \frac{\cos 2}{8} + \frac{1}{8} = \frac{3}{8} - \frac{\sin 2}{4} - \frac{\cos 2}{8}$$

Numerically:

$$I_2 \approx 0.375 - 0.227 - (-0.052) = 0.375 - 0.227 + 0.052 \approx 0.200$$

Calculating $I_1 = \int_0^1 x^2\sin^2 x \, dx$

Using $\sin^2 x = \frac{1 - \cos 2x}{2}$:

$$I_1 = \int_0^1 x^2 \cdot \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2}\int_0^1 x^2 \, dx - \frac{1}{2}\int_0^1 x^2\cos 2x \, dx$$

$$= \frac{1}{2} \cdot \frac{1}{3} - \frac{1}{2}\int_0^1 x^2\cos 2x \, dx = \frac{1}{6} - \frac{1}{2}\int_0^1 x^2\cos 2x \, dx$$

For $\int x^2\cos 2x \, dx$, use integration by parts twice:

First: $u = x^2$, $dv = \cos 2x \, dx$

$$\int x^2\cos 2x \, dx = \frac{x^2\sin 2x}{2} - \int x\sin 2x \, dx$$

Second: $u = x$, $dv = \sin 2x \, dx$

$$\int x\sin 2x \, dx = -\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2} \, dx = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4}$$

Therefore:

$$\int x^2\cos 2x \, dx = \frac{x^2\sin 2x}{2} + \frac{x\cos 2x}{2} - \frac{\sin 2x}{4}$$

Evaluating:

$$\int_0^1 x^2\cos 2x \, dx = \left[\frac{x^2\sin 2x}{2} + \frac{x\cos 2x}{2} - \frac{\sin 2x}{4}\right]_0^1$$

$$= \left(\frac{\sin 2}{2} + \frac{\cos 2}{2} - \frac{\sin 2}{4}\right) - 0 = \frac{\sin 2}{4} + \frac{\cos 2}{2}$$

Therefore:

$$I_1 = \frac{1}{6} - \frac{1}{2}\left(\frac{\sin 2}{4} + \frac{\cos 2}{2}\right) = \frac{1}{6} - \frac{\sin 2}{8} - \frac{\cos 2}{4}$$

Numerically:

$$I_1 \approx 0.167 - 0.114 - (-0.104) = 0.167 - 0.114 + 0.104 \approx 0.157$$

Verification

$$I_2 \approx 0.200 > I_1 \approx 0.157$$

This confirms: $I_2 > I_1$ ✓

Summary

Comparison by sign analysis:

$$I_1 - I_2 = \int_0^1 x(x-1)\sin^2 x \, dx < 0$$

Since $x(x-1) < 0$ for $x \in (0,1)$ and $\sin^2 x > 0$.

$$\boxed{\int_0^1 x\sin^2 x \, dx > \int_0^1 x^2\sin^2 x \, dx}$$

Numerical values:

$$I_2 \approx 0.200 > I_1 \approx 0.157$$

Answer: $I_2 = \int_0^1 x\sin^2 x \, dx$ is larger.

Key insights:

• On $(0,1)$: $x < x^2$ is FALSE (actually $x > x^2$ for $0 < x < 1$)
• The factor $(x-1) < 0$ makes $x(x-1) < 0$
• Direct comparison of integrands: $x\sin^2 x > x^2\sin^2 x$ for $x \in (0,1)$
• This is because $x > x^2$ when $0 < x < 1$
• The sign analysis method is elegant and avoids computation