Progressive Calculus practice questions
Use Clevolab for progressive practice in calculus. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats calculus as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current Progressive set covers differentiation, integration, rates of change, areas, and applications. 221 reviewed questions currently published for this page.
What you can practise
- Gradients and rates of change
- Differentiation rules
- Stationary points
- Integration basics
- Applied calculus questions
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
Find $\int 1\,dx$.
- A$x+C$Correct answer
- B$1+C$Answer option
- C$\ln x+C$Answer option
- D$x^2+C$Answer option
Why this answer is right
The antiderivative of $1$ is $x+C$, since $\dfrac{d}{dx}(x)=1$.
Because $\dfrac{d}{dx}(x)=1$, $$\int 1\,dx=x+C.$$ This captures the family of functions with constant slope $1$.
Sample question
Compute $\displaystyle \int 4x^3\,dx$.
- A$x^4+C$Correct answer
- B$4x^4+C$Answer option
- C$x^3+C$Answer option
- D$4x^3+C$Answer option
Why this answer is right
Use the reverse power rule: $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ for $n\neq -1$. Here $\int 4x^3dx=4\cdot\dfrac{x^{4}}{4}=x^4+C$.
Antidifferentiation reverses differentiation. For $x^n$ with $n\ne -1$, $$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C.$$
Sample question
For $|x|<1$, find a closed form for $\displaystyle S(x)=\sum_{n=1}^{\infty}\frac{n}{n+1}x^n$.
- A$\dfrac{x}{1-x}-\ln(1-x)$Answer option
- B$\dfrac{x}{1-x}+\dfrac{\ln(1-x)}{x}+1$Correct answer
- C$\dfrac{x}{(1-x)^2}-\ln(1-x)$Answer option
- D$\dfrac{x^2}{1-x}+\ln(1-x)$Answer option
Why this answer is right
Split $\dfrac{n}{n+1}=1-\dfrac{1}{n+1}$. Then $S=\sum x^n-\sum\dfrac{x^n}{n+1}=\dfrac{x}{1-x}-\dfrac{1}{x}\sum\dfrac{x^{n+1}}{n+1}$. Using $\sum\dfrac{x^m}{m}=-\ln(1-x)$ gives $S=\dfrac{x}{1-x}+\dfrac{\ln(1-x)}{x}+1$.
Write $$\frac{n}{n+1}=1-\frac{1}{n+1}.$$ Then $$S=\sum_{n=1}^{\infty}x^n-\sum_{n=1}^{\infty}\frac{x^n}{n+1}=\frac{x}{1-x}-\frac{1}{x}\sum_{m=2}^{\infty}\frac{x^{m}}{m}.$$
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.