Which is larger: I_a = integral from 1 to 4 a^x arctan x

Which is larger: $$ I_a = \int_1^4 a^x \arctan x \, dx \quad \text{or} \quad I_e = \int_1^4 e^x \arctan x \, dx $$

The answer depends on the value of $a$.

Analysis: Depends on $a$

Consider the difference:

$$I_e - I_a = \int_1^4 e^x \arctan x \, dx - \int_1^4 a^x \arctan x \, dx$$

$$= \int_1^4 (e^x - a^x)\arctan x \, dx$$

Step 1: Analyze $\arctan x$ on $[1,4]$

For $x \in [1,4]$:

$$\arctan x > 0$$

Specifically:

• $\arctan 1 = \frac{\pi}{4} \approx 0.785$
• $\arctan 4 \approx 1.326$

So $\arctan x \in \left[\frac{\pi}{4}, \arctan 4\right]$ and is always positive.

Step 2: Analyze the sign of $e^x - a^x$

The sign of $(e^x - a^x)$ depends on the relationship between $a$ and $e$.

Case 1: $a < e$

Then $a^x < e^x$ for all $x > 0$, so:

$$e^x - a^x > 0 \quad \text{for } x \in [1,4]$$

Therefore:

$$I_e - I_a = \int_1^4 (e^x - a^x)\arctan x \, dx > 0$$

$$\boxed{I_e > I_a \quad \text{when } a < e}$$

Case 2: $a = e$

Then $a^x = e^x$ for all $x$, so:

$$I_e - I_a = 0$$

$$\boxed{I_e = I_a \quad \text{when } a = e}$$

Case 3: $a > e$

Then $a^x > e^x$ for all $x > 0$, so:

$$e^x - a^x < 0 \quad \text{for } x \in [1,4]$$

Therefore:

$$I_e - I_a = \int_1^4 (e^x - a^x)\arctan x \, dx < 0$$

$$\boxed{I_e < I_a \quad \text{when } a > e}$$

Summary of comparison

$$\boxed{ \begin{cases} I_e > I_a & \text{if } a < e \approx 2.718 \\ I_e = I_a & \text{if } a = e \\ I_e < I_a & \text{if } a > e \end{cases} }$$

Special cases

Example 1: $a = 2 < e$

Since $2 < e$, we have $2^x < e^x$ for $x \in [1,4]$.

Therefore: $\int_1^4 e^x \arctan x \, dx > \int_1^4 2^x \arctan x \, dx$

Example 2: $a = 3 > e$

Since $3 > e$, we have $3^x > e^x$ for $x \in [1,4]$.

Therefore: $\int_1^4 e^x \arctan x \, dx < \int_1^4 3^x \arctan x \, dx$

Example 3: $a = \pi > e$

Since $\pi \approx 3.14159 > e$, we have $\pi^x > e^x$ for $x \in [1,4]$.

Therefore: $\int_1^4 e^x \arctan x \, dx < \int_1^4 \pi^x \arctan x \, dx$

Verification using specific values

Let's verify with $a = 2$:

For $x = 1$:

$$2^1 = 2 < e^1 \approx 2.718 \quad \checkmark$$

For $x = 4$:

$$2^4 = 16 < e^4 \approx 54.6 \quad \checkmark$$

Indeed, $2^x < e^x$ for all $x \in [1,4]$.

Let's verify with $a = 3$:

For $x = 1$:

$$3^1 = 3 > e^1 \approx 2.718 \quad \checkmark$$

For $x = 4$:

$$3^4 = 81 > e^4 \approx 54.6 \quad \checkmark$$

Indeed, $3^x > e^x$ for all $x \in [1,4]$.

General principle

For any positive function $f(x)$ (like $\arctan x$ on $[1,4]$):

$$\int_1^4 b^x f(x) \, dx \lessgtr \int_1^4 c^x f(x) \, dx$$

depending on whether $b \lessgtr c$.

Specifically, if $f(x) > 0$ on the interval:

$$b < c \implies \int_1^4 b^x f(x) \, dx < \int_1^4 c^x f(x) \, dx$$

Critical value

The critical value is $a = e \approx 2.71828$.

• For $a < e$: $I_e$ is larger
• For $a = e$: They are equal
• For $a > e$: $I_a$ is larger

Answer

$$\boxed{\text{The answer depends on } a:}$$

$$\boxed{ \begin{array}{lcl} a < e & \implies & \int_1^4 e^x \arctan x \, dx > \int_1^4 a^x \arctan x \, dx \\ a = e & \implies & \int_1^4 e^x \arctan x \, dx = \int_1^4 a^x \arctan x \, dx \\ a > e & \implies & \int_1^4 e^x \arctan x \, dx < \int_1^4 a^x \arctan x \, dx \end{array} }$$

Key insights:

• The comparison depends entirely on whether $a < e$, $a = e$, or $a > e$
• Since $\arctan x > 0$ on $[1,4]$, the sign of $(e^x - a^x)$ determines the result
• Exponential functions preserve order: if $b < c$ then $b^x < c^x$ for $x > 0$
• The critical threshold is $e \approx 2.718$
• Without knowing the value of $a$, we cannot definitively answer which is larger