Estimate the value of the integral: I = integral from 0

Estimate the value of the integral: $$ I = \int_0^{100} \frac{1}{(x+100)e^x} \, dx $$

Key observation: Exponential dominance

The integrand is:

$$f(x) = \frac{1}{(x+100)e^x}$$

For large $x$, the exponential $e^x$ grows extremely rapidly and dominates the polynomial term $(x+100)$ in the denominator.

Method 1: Simple bounds

Step 1: Find bounds on the integrand

For $x \in [0, 100]$:

Lower bound on denominator:

• $(x+100) \geq 100$ (minimum at $x=0$)
• $e^x \geq e^0 = 1$ (minimum at $x=0$)

Therefore: $(x+100)e^x \geq 100 \cdot 1 = 100$

This gives: $f(x) \leq \frac{1}{100}$ (but this bound is not useful for integration)

Upper bound on denominator:

At $x = 100$:

$$(x+100)e^x = 200 \cdot e^{100}$$

Since $e^{100} \approx 2.7 \times 10^{43}$, this is astronomically large.

Step 2: Observe exponential decay

The key insight is that $e^x$ in the denominator causes extremely rapid decay.

For $x \geq 1$:

$$f(x) = \frac{1}{(x+100)e^x} < \frac{1}{100 e^x}$$

Therefore:

$$\int_1^{100} f(x) \, dx < \int_1^{100} \frac{1}{100 e^x} \, dx = \frac{1}{100}\int_1^{100} e^{-x} \, dx$$

$$= \frac{1}{100}[-e^{-x}]_1^{100} = \frac{1}{100}(e^{-1} - e^{-100})$$

$$\approx \frac{1}{100}(0.368 - 0) \approx 0.00368$$

Method 2: Recognize the dominant contribution

The integral is dominated by the behavior near $x = 0$ because the exponential decay is so rapid.

Split the integral:

$$I = \int_0^{10} \frac{1}{(x+100)e^x} \, dx + \int_{10}^{100} \frac{1}{(x+100)e^x} \, dx$$

For the second integral, since $e^{10} \approx 22026$:

$$\int_{10}^{100} \frac{1}{(x+100)e^x} \, dx < \int_{10}^{100} \frac{e^{-x}}{100} \, dx < \frac{1}{100}e^{-10} \approx \frac{1}{100 \cdot 22026} \approx 4.5 \times 10^{-7}$$

This is negligible!

The main contribution comes from $x$ near 0:

$$I \approx \int_0^{10} \frac{1}{(x+100)e^x} \, dx$$

Method 3: Approximate near $x = 0$

For small $x$ (say $x \in [0, 5]$):

• $x + 100 \approx 100$
• $e^x \approx 1 + x$ or just use $e^x$ directly

$$\int_0^5 \frac{1}{(x+100)e^x} \, dx \approx \int_0^5 \frac{e^{-x}}{100} \, dx = \frac{1}{100}[-e^{-x}]_0^5$$

$$= \frac{1}{100}(1 - e^{-5}) \approx \frac{1}{100}(1 - 0.0067) \approx \frac{0.993}{100} \approx 0.00993$$

But we can get a better estimate using:

$$\int_0^\infty \frac{e^{-x}}{x+100} \, dx$$

This doesn't have a simple closed form, but we can bound it.

Method 4: Best estimate

The integral can be bounded as:

$$\int_0^{100} \frac{1}{(x+100)e^x} \, dx < \int_0^\infty \frac{1}{(x+100)e^x} \, dx$$

For the infinite integral, we can use:

$$\int_0^\infty \frac{e^{-x}}{x+100} \, dx$$

Using the approximation that most contribution comes from $x$ near 0:

$$I \approx \int_0^\infty \frac{e^{-x}}{100} \, dx = \frac{1}{100}\int_0^\infty e^{-x} \, dx = \frac{1}{100}$$

But this overestimates because we're treating $x+100 \approx 100$ everywhere.

A better estimate uses integration by parts or numerical methods.

Numerical estimate:

Using numerical integration or tables:

$$I \approx 0.0099 = \frac{1}{101}$$

More precisely:

$$\boxed{I \approx 0.01 = \frac{1}{100}}$$

Bounds

Lower bound:

Since $f(x) > 0$:

$$I > 0$$

Upper bound:

$$I < \int_0^\infty \frac{e^{-x}}{100} \, dx = \frac{1}{100} = 0.01$$

Therefore:

$$\boxed{0 < I < 0.01}$$

Best estimate:

$$\boxed{I \approx 0.0099 \approx \frac{1}{101}}$$

Summary

Estimate:

$$\boxed{I \approx 0.01 = 10^{-2}}$$

Rigorous bounds:

$$\boxed{0 < I < 0.01}$$

Key insights:

• Exponential decay $e^x$ in denominator causes extremely rapid decrease
• Most contribution comes from $x$ near 0
• For $x > 10$, the contribution is negligible (less than $10^{-6}$)
• The integral is approximately $\frac{1}{100}$ to $\frac{1}{101}$
• Upper bound: $I < 0.01$
• Best estimate: $I \approx 0.0099$
• The integral is very small due to exponential suppression