The calculus and linear algebra behind science, engineering and machine learning: derivatives, integrals, vectors, matrices, gradients and eigenvalues, with Python exercises.

Calculus & Linear Algebra for STEM

## Why this matters

The derivative answers "how fast?". The integral answers "how much in total?". If you know a pump's flow rate over time, the integral gives the total volume pumped. If you know the force along a spring, the integral gives the work done. In probability, the area under a density curve is a probability, and in pharmacology the area under a concentration curve (AUC) measures total drug exposure.

## Area as accumulation

When a rate is constant, total = rate × time: 3 litres per minute for 10 minutes is 30 litres, the area of a 3-by-10 rectangle. When the rate changes, the total is still the **area under the rate curve**. The integral is the tool for computing that area.

The **definite integral** of f from a to b is written:

```text
∫ from a to b of f(x) dx
```

The long S stands for "sum", f(x) is the height, and dx is a tiny width. You are adding up many thin rectangles.

## Riemann sums

To approximate the area, chop [a, b] into n strips of width Δx = (b − a) / n, build a rectangle on each one, and add up their areas. The height can come from the left edge, the right edge or the midpoint of each strip:

```text
left sum     = Δx · [f(x₀) + f(x₁) + … + f(xₙ₋₁)]
midpoint sum = Δx · Σ f(a + (i + 0.5)·Δx),  i = 0 … n − 1
```

Worked example: f(x) = x² on [0, 3] with n = 3 strips (Δx = 1).

| strip | midpoint | height | area |
|---|---|---|---|
| [0, 1] | 0.5 | 0.25 | 0.25 |
| [1, 2] | 1.5 | 2.25 | 2.25 |
| [2, 3] | 2.5 | 6.25 | 6.25 |

The midpoint sum is 8.75. With more strips it approaches the exact value, 9. That limit, as the strips get infinitely thin, *is* the definite integral.

## The fundamental theorem of calculus

Adding up rectangles works, but there is a shortcut that links integrals back to derivatives. If F is any function whose derivative is f (an **antiderivative**), then:

```text
∫ from a to b of f(x) dx = F(b) − F(a)
```

For f(x) = x², an antiderivative is F(x) = x³/3, so the area on [0, 3] is 27/3 − 0 = 9. This is the **fundamental theorem of calculus**: accumulation and rate of change are inverse operations. Reversing the power rule gives the most common antiderivative:

```text
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C      (n ≠ −1)
```

The + C appears in an *indefinite* integral (no limits) because any constant disappears when differentiated. It cancels in a definite integral.

## Numerical integration in practice

Lab instruments and simulations produce data points, not formulas, so numerical integration is routine. The **trapezoidal rule** joins neighbouring points with straight lines; NumPy has it built in as `numpy.trapezoid` (`numpy.trapz` in older versions), and SciPy offers `scipy.integrate.quad` for formulas. A plain Python midpoint rule takes a few lines:

```python
def midpoint_integral(f, a, b, n):
    dx = (b - a) / n
    return dx * sum(f(a + (i + 0.5) * dx) for i in range(n))

print(midpoint_integral(lambda x: x ** 2, 0, 3, 1000))   # about 8.99999775
```

## Signed area

Area below the x-axis counts as negative. The integral of sin x from 0 to 2π is 0: the positive hump and the negative hump cancel. If you want total distance rather than net displacement, integrate the absolute value of the speed.

## Common mistakes

- **Too few strips.** Accuracy improves as n grows. Compare n and 2n; if the answers differ noticeably, use more.
- **Mixing up Δx.** The strip width is (b − a) / n, not 1 / n.
- **Forgetting units.** The integral of a flow rate in L/min over minutes is litres: multiply the units, just as with rectangles.
- **Dropping + C** on an indefinite integral, or keeping it in a definite one.

## Quick recap

- A definite integral is the limit of a sum of thin rectangles: area or total accumulation.
- Riemann sums (left, right, midpoint) approximate it; more strips means a better answer.
- The fundamental theorem: ∫ from a to b of f dx = F(b) − F(a), where F′ = f.
- Area below the axis counts as negative.

## Try it

In the **Do** tab, write the midpoint Riemann sum and watch it converge to the exact area.


Treat the integral as accumulated area, approximate it with Riemann sums and connect it to derivatives.

Area and the integral

Riemann approximation introduction | Accumulation and Riemann sums | AP Calculus AB | Khan Academy

Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

## Why this matters

The fundamental theorem says that to integrate f, find a function whose derivative is f. For powers that is easy, but many integrals you meet in physics, chemistry and statistics hide a function inside another function. **Substitution** undoes the chain rule and handles most of them. You will rarely integrate complicated expressions by hand at work, since tools like SymPy, Wolfram Alpha and numerical integration do it, but recognising the patterns helps you set up models and check results.

## A table of basic antiderivatives

| f(x) | ∫ f(x) dx |
|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹ / (n + 1) + C |
| 1/x | ln\|x\| + C |
| eˣ | eˣ + C |
| e^(kx) | e^(kx) / k + C |
| sin x | −cos x + C |
| cos x | sin x + C |

Every entry can be checked by differentiating the right-hand side. That habit, differentiate your answer, catches almost every integration error.

## The core idea of substitution

The chain rule says the derivative of F(g(x)) is F′(g(x)) · g′(x). Read backwards: if an integral has the form "a function of g(x), times g′(x)", substitute u = g(x) and du = g′(x) dx, and it collapses to a simple integral in u.

The steps:

1. Choose u to be the "inside" expression, often the part inside parentheses, a root, or an exponent.
2. Compute du = u′(x) dx.
3. Rewrite the whole integral in terms of u. Nothing with x should remain.
4. Integrate in u.
5. Substitute back u = g(x).

## Worked example 1

```text
∫ 2x (x² + 1)⁵ dx
```

Let u = x² + 1. Then du = 2x dx, which is exactly the factor in front.

```text
∫ u⁵ du = u⁶ / 6 + C = (x² + 1)⁶ / 6 + C
```

Check: the derivative of (x² + 1)⁶ / 6 is 6(x² + 1)⁵ · 2x / 6 = 2x(x² + 1)⁵. Correct.

## Worked example 2: fixing a constant

```text
∫ x · e^(3x²) dx
```

Let u = 3x², so du = 6x dx and x dx = du / 6.

```text
∫ e^u · (du / 6) = e^u / 6 + C = e^(3x²) / 6 + C
```

Constant factors are easy to fix. If the leftover piece involved x itself (say x² instead of x), substitution with this u would not work and you would try a different u or a different method.

## Definite integrals: change the limits

With limits, convert them to u-values as well and you never need to substitute back:

```text
∫ from 0 to 1 of 2x (x² + 1)³ dx,   u = x² + 1
x = 0 → u = 1,   x = 1 → u = 2
∫ from 1 to 2 of u³ du = [u⁴ / 4] from 1 to 2 = 4 − 1/4 = 3.75
```

A midpoint sum in code gives the same 3.75, a quick way to confirm.

## Other techniques, briefly

- **Integration by parts** reverses the product rule: ∫ u dv = uv − ∫ v du. It handles products like x·eˣ or x·ln x.
- **Partial fractions** split rational functions such as 1 / ((x − 1)(x + 2)) into simpler pieces; they appear in reaction kinetics and control systems.
- **Numerical integration** (trapezoid, Simpson's rule, `scipy.integrate.quad`) is the workhorse when no closed form exists, for example the bell curve e^(−x²), whose antiderivative cannot be written with elementary functions. That is why normal-distribution probabilities come from tables or software.

## Common mistakes

- **Leaving some x behind.** After substituting, the integral must be entirely in u and du.
- **Forgetting du.** Replacing g(x) with u but not dx with du / g′(x) gives the wrong constant.
- **Keeping the old limits** after switching to u in a definite integral.
- **Skipping the check.** Differentiating your answer takes seconds.

## Quick recap

- Know the basic antiderivatives; check any answer by differentiating it.
- Substitution reverses the chain rule: pick the inside function as u, replace dx using du.
- In definite integrals, convert the limits to u.
- Integration by parts, partial fractions and numerical methods cover most of the rest.

## Try it

This lesson has no code exercise. Work through the quiz in the **Do** tab, ideally with pencil and paper.


Use a table of antiderivatives and u-substitution to integrate functions that hide a chain rule.

Integration techniques

How To Integrate Using U-Substitution

## Why this matters

Many laws of nature are stated as rates. A radioactive sample decays at a rate proportional to how much is left. A cup of coffee cools at a rate proportional to its temperature difference with the room. A population grows in proportion to its size until resources run out. Equations that relate a quantity to its own rate of change are **differential equations**, and they are the language of physics, chemistry, epidemiology, finance and control engineering.

## The core idea

A differential equation gives a rule for the derivative. The simplest important one is:

```text
dy/dt = k · y
```

"The rate of change of y is proportional to y." Its solution is exponential:

```text
y(t) = y₀ · e^(kt)
```

You can check it: the derivative of y₀e^(kt) is k · y₀e^(kt) = k · y. With k > 0 it describes growth (compound interest, bacteria); with k < 0 it describes decay (radioactivity, drug elimination, a capacitor discharging).

Worked example: a drug is eliminated at 30% of the current amount per hour (k = −0.3) and the starting dose is 200 mg. After 4 hours: y = 200 · e^(−1.2) ≈ 60.2 mg. The **half-life** is ln 2 / 0.3 ≈ 2.31 hours.

## Initial conditions

A differential equation alone describes a family of curves. The **initial condition** (y₀ at t = 0) picks the one that matches your situation, just as the + C in an indefinite integral is fixed by a known value.

## When there is no formula: Euler's method

Most real models, such as the logistic growth equation dy/dt = r·y·(1 − y/K), predator-prey systems or a pendulum with friction, are solved numerically. **Euler's method** is the simplest approach and the idea behind every solver:

1. Start at (t₀, y₀).
2. The equation tells you the slope there: f(t₀, y₀).
3. Take a small step of size h along that slope: y₁ = y₀ + h · f(t₀, y₀), t₁ = t₀ + h.
4. Repeat from the new point.

```python
def euler(f, y0, t0, t1, n):
    h = (t1 - t0) / n
    t, y = t0, y0
    for _ in range(n):
        y = y + h * f(t, y)
        t = t + h
    return y

decay = lambda t, y: -0.3 * y
print(euler(decay, 200, 0, 4, 1000))   # about 60.2, close to 200 * e^(-1.2)
```

Worked by hand with only two steps (h = 2): y₁ = 200 + 2 · (−60) = 80, then y₂ = 80 + 2 · (−24) = 32. That is far from 60.2. Euler's method is only accurate when the steps are small, because it follows a straight line while the true curve bends.

## Professional solvers

In practice you would call a library: `scipy.integrate.solve_ivp` uses Runge-Kutta methods, which sample the slope several times per step for much better accuracy, and adapts the step size automatically. MATLAB's `ode45` is the same family. Knowing Euler's method tells you what these tools are doing and why step size matters.

## Common mistakes

- **Steps that are too large.** Halve h and compare. If the answer changes a lot, the step is too big. With large steps a decay model can even oscillate or blow up.
- **Mixing up the rate constant's sign.** Decay needs k < 0 in dy/dt = ky (or write −k with k > 0).
- **Forgetting units.** If t is in hours, k is per hour.
- **Ignoring the initial condition.** Without y₀ there is no unique answer.

## Quick recap

- A differential equation relates a quantity to its rate of change.
- dy/dt = ky has the exponential solution y₀e^(kt): growth for k > 0, decay for k < 0.
- Euler's method steps along the slope: y ← y + h · f(t, y).
- Smaller steps are more accurate; professional solvers use smarter, adaptive methods.

## Try it

In the **Do** tab, write the Euler update and simulate decay and growth.


Model growth and decay with dy/dt = ky, and simulate equations step by step with Euler's method.

Intro to differential equations

Euler's method | Differential equations| AP Calculus BC | Khan Academy

Differential equations, a tourist's guide | DE1

Integrals as accumulated area, the fundamental theorem, substitution, and differential equations solved numerically.