Gilbert Strang’s Calculus: Big Picture Integral

Calculus

Integrals

Derivatives

Author

Chao Ma

Published

February 14, 2026

Integrals and derivatives can be mostly explained by working (very briefly) with sums and differences.
- Gilbert Strang

The connection between two functions

The key connection is between a height/distance function and a slope function. This pair is a two-way map: derivative gives local change from total accumulation, and integral rebuilds total accumulation from local change (up to an initial value).

Height/Distance function: $y(x)$
This function represents accumulated height or distance.
Slope function: \[ s(x)=\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} \] This function represents the instantaneous rate of change (for distance-time, this is speed).

Examples:

If $y=x^n$, then $\frac{dy}{dx}=nx^{n-1}$.
If $\frac{dy}{dx}=x^n$, then \[ y(x)=\frac{x^{n+1}}{n+1}+C. \]

From slope to distance

Suppose we know the slope each hour and want distance back.

Hourly slopes: \[ 4,\ 3,\ 2,\ 1,\ 0. \]

Cumulative distance:

Hour 1: $4$
Hour 2: $4+3=7$
Hour 3: $7+2=9$
Hour 4: $9+1=10$
Hour 5: $10+0=10$

From an algebra view, sample distance at times as $y_0,y_1,y_2,\dots,y_n$. Then each increment is: \[ \Delta y_k=y_k-y_{k-1}. \] Summing all increments gives: \[ \sum_{k=1}^n \Delta y_k =(y_1-y_0)+(y_2-y_1)+\cdots+(y_n-y_{n-1}) =y_n-y_0. \]

This is exactly “final value minus initial value.”

Another equivalent form is: \[ \sum \left(\frac{\Delta y}{\Delta x}\right)\Delta x = y_{\text{last}}-y_{\text{first}}. \]

As $\Delta x$ gets smaller and smaller, $\frac{\Delta y}{\Delta x}$ approaches the slope function $\frac{dy}{dx}$.

The answer is as simple as changing $\sum$ to $\int$:

\[ \int s(x)\,dx. \]

Here $s(x)$ is the slope, and $dx$ is an infinitesimal change in $x$. Geometrically, this is accumulated area under the slope curve.

Example

Slope function: \[ s(x)=2-2x. \]

This starts at 2 (when $x=0$) and drops to 0 (when $x=1$). To get height/distance, we can first look at a discrete approximation, then the integral limit.

Discrete point of view

Split $[0,1]$ into 4 equal intervals, so $\Delta x=\frac14$. Use left endpoints as piecewise-constant speeds.

Speeds at left endpoints:

$x=0$: $2$
$x=\frac14$: $1.5$
$x=\frac12$: $1$
$x=\frac34$: $0.5$

Left Riemann sum: \[ 2\Delta x+1.5\Delta x+1\Delta x+0.5\Delta x =5\Delta x=\frac54=1.25. \]

Observation:

Inside each interval, slope is decreasing.
Using the left endpoint gives a larger rectangle than the true area.
So this is an overestimate.

Now use smaller intervals, for example $\Delta x=\frac18$: \[ 2\Delta x+1.75\Delta x+1.5\Delta x+1.25\Delta x+1\Delta x+0.75\Delta x+0.5\Delta x+0.25\Delta x =9\Delta x=\frac98=1.125. \]

The estimate gets closer as $\Delta x\to 0$.

Integral point of view

From the discrete view, smaller $\Delta x$ gives better accuracy. In the limit, rectangles become infinitesimally thin, and the sum becomes the exact area: \[ \int_0^x (2-2t)\,dt. \]

Areas:

At $x=1$: area is triangle area \[ \frac{1\cdot 2}{2}=1. \]
At $x=0.5$: area can be computed either as
- big triangle minus small triangle, or
- trapezoid area from $x=0$ to $x=0.5$.

Both give $0.75$.

Proof

Find an antiderivative of $s(x)=2-2x$: \[ y(x)=2x-x^2+C. \] If we set $y(0)=0$, then $C=0$, so \[ y(x)=2x-x^2. \]

Check:

$x=1$: $y(1)=2(1)-1^2=1$
$x=0.5$: $y(0.5)=2(0.5)-(0.5)^2=1-0.25=0.75$

So the integral view and algebraic antiderivative match exactly.

--- title: "Gilbert Strang's Calculus: Big Picture Integral" author: "Chao Ma" date: "2026-02-14" categories: [Calculus, Integrals, Derivatives] --- {{< video https://www.youtube.com/watch?v=2qxY859dzzQ&list=PLFW_V3qDH5jRyfpD9uiq6aKVTWIxpbIm3&index=5 >}} Integrals and derivatives can be mostly explained by working (very briefly) with sums and differences. - Gilbert Strang # The connection between two functions The key connection is between a height/distance function and a slope function. This pair is a two-way map: derivative gives local change from total accumulation, and integral rebuilds total accumulation from local change (up to an initial value). - Height/Distance function: $y(x)$ This function represents accumulated height or distance. - Slope function: $$ s(x)=\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} $$ This function represents the instantaneous rate of change (for distance-time, this is speed). Examples: - If $y=x^n$, then $\frac{dy}{dx}=nx^{n-1}$. - If $\frac{dy}{dx}=x^n$, then $$ y(x)=\frac{x^{n+1}}{n+1}+C. $$ # From slope to distance Suppose we know the slope each hour and want distance back. Hourly slopes: $$ 4,\ 3,\ 2,\ 1,\ 0. $$ Cumulative distance: - Hour 1: $4$ - Hour 2: $4+3=7$ - Hour 3: $7+2=9$ - Hour 4: $9+1=10$ - Hour 5: $10+0=10$ From an algebra view, sample distance at times as $y_0,y_1,y_2,\dots,y_n$. Then each increment is: $$ \Delta y_k=y_k-y_{k-1}. $$ Summing all increments gives: $$ \sum_{k=1}^n \Delta y_k =(y_1-y_0)+(y_2-y_1)+\cdots+(y_n-y_{n-1}) =y_n-y_0. $$ This is exactly “final value minus initial value.” Another equivalent form is: $$ \sum \left(\frac{\Delta y}{\Delta x}\right)\Delta x = y_{\text{last}}-y_{\text{first}}. $$ As $\Delta x$ gets smaller and smaller, $\frac{\Delta y}{\Delta x}$ approaches the slope function $\frac{dy}{dx}$. The answer is as simple as changing $\sum$ to $\int$: $$ \int s(x)\,dx. $$ Here $s(x)$ is the slope, and $dx$ is an infinitesimal change in $x$. Geometrically, this is accumulated area under the slope curve. # Example Slope function: $$ s(x)=2-2x. $$ This starts at 2 (when $x=0$) and drops to 0 (when $x=1$). To get height/distance, we can first look at a discrete approximation, then the integral limit. ## Discrete point of view Split $[0,1]$ into 4 equal intervals, so $\Delta x=\frac14$. Use left endpoints as piecewise-constant speeds. Speeds at left endpoints: - $x=0$: $2$ - $x=\frac14$: $1.5$ - $x=\frac12$: $1$ - $x=\frac34$: $0.5$ Left Riemann sum: $$ 2\Delta x+1.5\Delta x+1\Delta x+0.5\Delta x =5\Delta x=\frac54=1.25. $$ ![Discrete approximation with rectangles](big-picture-integral-1.png) Observation: - Inside each interval, slope is decreasing. - Using the left endpoint gives a larger rectangle than the true area. - So this is an overestimate. Now use smaller intervals, for example $\Delta x=\frac18$: $$ 2\Delta x+1.75\Delta x+1.5\Delta x+1.25\Delta x+1\Delta x+0.75\Delta x+0.5\Delta x+0.25\Delta x =9\Delta x=\frac98=1.125. $$ The estimate gets closer as $\Delta x\to 0$. ## Integral point of view From the discrete view, smaller $\Delta x$ gives better accuracy. In the limit, rectangles become infinitesimally thin, and the sum becomes the exact area: $$ \int_0^x (2-2t)\,dt. $$ ![Area interpretation in the limit](big-picture-integral-2.png) Areas: - At $x=1$: area is triangle area $$ \frac{1\cdot 2}{2}=1. $$ - At $x=0.5$: area can be computed either as - big triangle minus small triangle, or - trapezoid area from $x=0$ to $x=0.5$. Both give $0.75$. ## Proof Find an antiderivative of $s(x)=2-2x$: $$ y(x)=2x-x^2+C. $$ If we set $y(0)=0$, then $C=0$, so $$ y(x)=2x-x^2. $$ Check: - $x=1$: $y(1)=2(1)-1^2=1$ - $x=0.5$: $y(0.5)=2(0.5)-(0.5)^2=1-0.25=0.75$ So the integral view and algebraic antiderivative match exactly.