๐ Worked Example: Related Rates โ Expanding Circle
Problem:
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this The Integral Test study guide free?โพ
Yes โ all study notes, flashcards, and practice problems for The Integral Test on Study Mondo are 100% free. No account is needed to access the content.
What course covers The Integral Test?โพ
The Integral Test is part of the AP Calculus BC course on Study Mondo, specifically in the Sequences & Series (BC) section. You can explore the full course for more related topics and practice resources.
Are there practice problems for The Integral Test?
f(x)
Continuous for xโฅ1
Positive: f(x)>0
Decreasing: f(x+1)<f(x)
Then the series โn=1โโf(n) and the integral โซ1โโf(x)dx:
Both converge or both diverge together!
Visual Understanding
Imagine rectangles with:
Width: 1
Heights: f(1),f(2),f(3),โฆ
The sum โf(n) is the total area of rectangles.
The integral โซf(x)dx is the area under the curve.
Since f is decreasing:
Rectangles (left endpoints) > curve area
Rectangles (right endpoints) < curve area
So: โซ1โโf(x)dx<โn=1โโf(n)<f(1)+โซ1โโf(x)dx
How to Use the Integral Test
Step 1: Verify f(x) is continuous, positive, and decreasing for xโฅN (some starting point)
\text{converges} & \text{if } p > 1 \\
\text{diverges} & \text{if } p \leq 1
\end{cases}$$
> **๐ฏ MEMORIZE THIS!** Most commonly tested series!
---
## Proof Using Integral Test
**Case 1**: $p > 1$
$$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b \to \infty} \left[\frac{x^{-p+1}}{-p+1}\right]_1^b$$
$$= \lim_{b \to \infty} \left[\frac{1}{(1-p)x^{p-1}}\right]_1^b$$
$$= \frac{1}{1-p}\lim_{b \to \infty} \left(\frac{1}{b^{p-1}} - 1\right)$$
Since $p > 1$, we have $p - 1 > 0$, so $\frac{1}{b^{p-1}} \to 0$.
$$= \frac{1}{1-p}(0 - 1) = \frac{1}{p-1}$$ (finite!)
**Converges when $p > 1$.**
---
**Case 2**: $p = 1$
This is the harmonic series (we showed it diverges).
---
**Case 3**: $p < 1$
$$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \infty$$
**Diverges when $p < 1$.**
---
## Example 2: Test $\sum_{n=1}^{\infty} \frac{1}{n^2}$
This is a p-series with $p = 2 > 1$.
**By p-series test: converges!**
(Actually converges to $\frac{\pi^2}{6}$, discovered by Euler!)
---
## Example 3: Test $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
Rewrite: $\sum \frac{1}{n^{1/2}}$
This is a p-series with $p = \frac{1}{2} < 1$.
**By p-series test: diverges!**
---
## Example 4: Use Integral Test
Test $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ for convergence.
**Step 1: Check conditions**
$f(x) = \frac{x}{x^2 + 1}$
โ Continuous for $x \geq 1$
โ Positive for $x > 0$
Check decreasing: $f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$
For $x > 1$: $1 - x^2 < 0$, so $f'(x) < 0$ โ
---
**Step 2: Evaluate integral**
$$\int_1^{\infty} \frac{x}{x^2+1}\,dx$$
**Use substitution**: $u = x^2 + 1$, $du = 2x\,dx$
$$= \frac{1}{2}\int_2^{\infty} \frac{1}{u}\,du = \frac{1}{2}\lim_{b \to \infty} [\ln u]_2^b$$
$$= \frac{1}{2}\lim_{b \to \infty} (\ln b - \ln 2) = \infty$$
---
**Step 3: Conclusion**
The integral diverges, so **the series diverges**.
---
## Example 5: Test $\sum_{n=1}^{\infty} \frac{\ln n}{n}$
**Step 1: Check conditions** (for $n \geq 2$ to avoid $\ln 1 = 0$)
$f(x) = \frac{\ln x}{x}$ for $x \geq 2$
โ Continuous
โ Positive for $x > 1$
Check decreasing: Use quotient rule or note that for large $x$, denominator grows faster.
$f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$
For $x > e \approx 2.718$: $\ln x > 1$, so $f'(x) < 0$ โ
---
**Step 2: Evaluate integral**
$$\int_2^{\infty} \frac{\ln x}{x}\,dx$$
**Use substitution**: $u = \ln x$, $du = \frac{1}{x}\,dx$
When $x = 2$: $u = \ln 2$
When $x \to \infty$: $u \to \infty$
$$= \int_{\ln 2}^{\infty} u\,du = \lim_{b \to \infty} \left[\frac{u^2}{2}\right]_{\ln 2}^b = \infty$$
---
**Step 3: Conclusion**
**The series diverges.**
---
## Important Notes About Integral Test
### Note 1: Value Not Equal
The Integral Test tells you about **convergence**, not the **value**!
$$\sum_{n=1}^{\infty} f(n) \neq \int_1^{\infty} f(x)\,dx$$
(They're close, but not equal)
---
### Note 2: Can Start Anywhere
You can start the series and integral at any point $N$:
$$\sum_{n=N}^{\infty} f(n) \text{ and } \int_N^{\infty} f(x)\,dx$$
Both converge or both diverge (though values differ).
---
### Note 3: Must Be Decreasing
If $f(x)$ is not eventually decreasing, can't use Integral Test!
**Example**: $f(x) = \sin x$ oscillates, not decreasing.
---
### Note 4: Remainder Estimate
If the series converges, the **remainder** (error after $n$ terms) satisfies:
$$\int_{n+1}^{\infty} f(x)\,dx < R_n < \int_n^{\infty} f(x)\,dx$$
where $R_n = \sum_{k=n+1}^{\infty} f(k)$
---
## Example 6: Estimate Error
For $\sum_{n=1}^{\infty} \frac{1}{n^2}$, estimate the error if we use the first 10 terms.
**Remainder is**:
$$R_{10} = \sum_{n=11}^{\infty} \frac{1}{n^2}$$
---
**Upper bound**:
$$R_{10} < \int_{10}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{10}^{\infty} = 0 - \left(-\frac{1}{10}\right) = \frac{1}{10} = 0.1$$
**Lower bound**:
$$R_{10} > \int_{11}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{11}^{\infty} = \frac{1}{11} \approx 0.091$$
---
**Conclusion**: $0.091 < R_{10} < 0.1$
The error is less than 0.1!
---
## โ ๏ธ Common Mistakes
### Mistake 1: Using When Not Decreasing
**WRONG**: Apply Integral Test to $\sum \frac{\sin n}{n}$
$f(x) = \frac{\sin x}{x}$ is not always decreasing (oscillates)!
---
### Mistake 2: Thinking Value Equals Integral
**WRONG**: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \int_1^{\infty} \frac{1}{x^2}\,dx = 1$
**RIGHT**: The series converges (by Integral Test), but its value is $\frac{\pi^2}{6} \approx 1.645$, not 1.
---
### Mistake 3: Forgetting p-Series Test
Don't do complicated integral when you can use p-series test directly!
For $\sum \frac{1}{n^p}$: just check if $p > 1$ (converges) or $p \leq 1$ (diverges).
---
### Mistake 4: Wrong Limits
Make sure integral limits match series starting point!
If series starts at $n = 2$, use $\int_2^{\infty}$.
---
## ๐ Practice Strategy
1. **Check p-series first**: Is it $\sum \frac{1}{n^p}$? Use p-series test!
2. **Verify conditions**: continuous, positive, decreasing (check derivative)
3. **Integrate**: Use substitution, partial fractions, etc.
4. **Improper integral**: Always use limits, check if finite or infinite
5. **For estimates**: Use remainder bounds $\int_{n+1}^{\infty} < R_n < \int_n^{\infty}$
6. **Memorize p-series**: $p > 1$ converges, $p \leq 1$ diverges
7. **When stuck**: Try to compare to p-series instead of integrating!
nlnn1โ
๐ก Show Solution
Step 1: Check conditions
f(x)=xlnx1โ for xโฅ2
โ Continuous for xโฅ2 (since lnx>0 for x>1)
โ Positive for x
Check decreasing: As x increases, both x and lnx increase, so xlnx decreases โ
Step 2: Evaluate integral
โซ2โโxlnx1โ
Use substitution: u=lnx, du=x1โd
When x=2: u=ln2
When xโโ: uโ
=โซln2โโ
=limbโโโ(lnbโln(ln2))=โ
Step 3: Conclusion
The integral diverges.
By the Integral Test, the series diverges.
Note: Even though terms go to 0 (slowly), they don't go to 0 fast enough!
2Problem 2easy
โ Question:
Determine if โn=1โโn32n+1โ converges or diverges.
๐ก Show Solution
Step 1: Split the series
โn=1โโ
3Problem 3hard
โ Question:
Use the Integral Test to show that โn=1โโneโn2 converges.
๐ก Show Solution
Step 1: Check conditions
f(x)=xeโx2 for
โพ
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
<
0
>
1
1
โ
d
x
x
โ
u
1
โ
d
u
=
limbโโโ[lnโฃuโฃ]ln2bโ
n32n+1โ
=
โn=1โโn32nโ+
โn=1โโn31โ
=โn=1โโn22โ+โn=1โโn31โ
=2โn=1โโn21โ+โn=1โโn31โ
Step 2: Identify p-series
โn21โ: p-series with p=2>1 โ converges
โn31โ: p-series with p=3>1 โ converges
Step 3: Use series properties
Sum of two convergent series converges.
The series converges.
Alternative Method (Integral Test):
Could also note that for large n: n32n+1โโn32nโ=n22โ
This behaves like n21โ (p-series with p=2>1), so converges.