Which is larger: I_1 = integral from 1 to ² 3^(x²) dx

Which is larger: $$ I_1 = \int_1^2 3^{x^2} \, dx \quad \text{or} \quad I_2 = \int_1^2 3^{x^3} \, dx $$

Method: Direct comparison

Consider the difference:

$$I_2 - I_1 = \int_1^2 3^{x^3} \, dx - \int_1^2 3^{x^2} \, dx = \int_1^2 (3^{x^3} - 3^{x^2}) \, dx$$

$$= \int_1^2 3^{x^2}(3^{x^3-x^2} - 1) \, dx$$

Step 1: Analyze the exponents $x^3$ and $x^2$ on $[1,2]$

We need to compare $x^3$ and $x^2$ for $x \in [1,2]$.

$$x^3 - x^2 = x^2(x - 1)$$

For $x \in [1,2]$:

• $x^2 > 0$
• $x - 1 \geq 0$ (equals 0 at $x=1$, positive for $x \in (1,2]$)

Therefore:

$$x^3 - x^2 = x^2(x-1) \geq 0 \quad \text{for } x \in [1,2]$$

Moreover, $x^3 - x^2 > 0$ for $x \in (1,2]$.

Step 2: Specific values

At $x = 1$:

$$x^3 - x^2 = 1 - 1 = 0$$

At $x = 1.5$:

$$x^3 - x^2 = 1.5^3 - 1.5^2 = 3.375 - 2.25 = 1.125 > 0$$

At $x = 2$:

$$x^3 - x^2 = 8 - 4 = 4 > 0$$

Step 3: Compare $3^{x^3}$ and $3^{x^2}$

Since $3 > 1$ and exponential functions are increasing:

For $x \in (1,2]$: $x^3 > x^2 \implies 3^{x^3} > 3^{x^2}$

At $x = 1$: $x^3 = x^2 \implies 3^{x^3} = 3^{x^2}$

Therefore:

$$3^{x^3} - 3^{x^2} \geq 0 \quad \text{for } x \in [1,2]$$

with strict inequality for $x \in (1,2]$.

Step 4: Determine the sign of the integral

$$I_2 - I_1 = \int_1^2 (3^{x^3} - 3^{x^2}) \, dx > 0$$

The integrand is non-negative on $[1,2]$ and strictly positive on $(1,2]$, so the integral is positive.

Therefore:

$$I_2 > I_1$$

$$\boxed{\int_1^2 3^{x^3} \, dx > \int_1^2 3^{x^2} \, dx}$$

Alternative explanation

For $x \in [1,2]$:

$$x \geq 1 \implies x^3 \geq x^2$$

More precisely, for $x > 1$:

$$x^3 > x^2 \iff x^2 \cdot x > x^2 \cdot 1 \iff x > 1 \quad \checkmark$$

Since the base $3 > 1$, the exponential function $3^t$ is strictly increasing in $t$.

Therefore:

$$x^3 \geq x^2 \implies 3^{x^3} \geq 3^{x^2}$$

with strict inequality for $x > 1$.

Integrating over $[1,2]$:

$$\int_1^2 3^{x^3} \, dx > \int_1^2 3^{x^2} \, dx$$

Numerical verification

Let's estimate the values:

At $x = 1$:

$$3^{1^2} = 3^1 = 3$$

$$3^{1^3} = 3^1 = 3$$

At $x = 2$:

$$3^{2^2} = 3^4 = 81$$

$$3^{2^3} = 3^8 = 6561$$

The integrand $3^{x^3}$ grows much faster than $3^{x^2}$ as $x$ increases from 1 to 2.

A rough estimate using the trapezoidal rule:

$$I_1 \approx \frac{3 + 81}{2} \cdot 1 = 42$$

$$I_2 \approx \frac{3 + 6561}{2} \cdot 1 = 3282$$

Clearly $I_2 \gg I_1$.

Summary

Comparison:

For $x \in [1,2]$:

$$x \geq 1 \implies x^3 \geq x^2 \implies 3^{x^3} \geq 3^{x^2}$$

with strict inequality for $x \in (1,2]$.

Therefore:

$$\boxed{\int_1^2 3^{x^3} \, dx > \int_1^2 3^{x^2} \, dx}$$

Answer: $I_2 = \int_1^2 3^{x^3} dx$ is larger.

Key insights:

• For $x > 1$: $x^3 > x^2$ (higher power dominates)
• For base $b > 1$: $b^t$ is increasing, so $t_1 > t_2 \implies b^{t_1} > b^{t_2}$
• The difference $x^3 - x^2 = x^2(x-1)$ is positive for $x > 1$
• At $x = 2$: $3^8 = 6561 \gg 3^4 = 81$ shows dramatic difference
• The integral $I_2$ is significantly larger than $I_1$