Integration and Accumulation of Change

Differentiation found rates. Integration runs the idea in reverse: given a rate of change, it finds the accumulated change 累积变化. The key picture: the area between the graph of a rate function and the $x$-axis gives the total accumulation.

• If the rate is positive over an interval, the accumulated change is positive; if negative, negative. Area below the axis counts as negative.
• Simple regions (triangles, rectangles) can be found with geometry 几何.
• Units: the area's unit is the rate's unit times the input's unit. A rate in vehicles-per-hour times hours gives vehicles.

When exact area is hard, approximate it with a Riemann sum 黎曼和 – split the interval into subintervals and add up rectangle (or trapezoid) areas. The four standard estimates:

Each rectangle has area $f(x)\,\Delta x$; adding them estimates the area, and as the strips narrow the sum approaches the definite integral.

• Left Riemann sum – rectangle height from the left endpoint of each subinterval.
• Right Riemann sum – height from the right endpoint.
• Midpoint Riemann sum – height from the midpoint 中点.
• Trapezoidal sum 梯形法 – average the two endpoint heights (a trapezoid).

Subintervals may be uniform (equal width) or nonuniform – read widths from the table.

Over- or underestimate? Judge from the behavior of the function: for an increasing function, a left sum underestimates and a right sum overestimates; a trapezoidal sum overestimates when the function is concave up and underestimates when concave down. Exam parts ask you to state which and why.

Worked example. A table gives $f(0)=3$, $f(2)=5$, $f(4)=8$, $f(6)=9$. Estimate $\int_0^6 f(x)\,dx$ with a right Riemann sum of three equal subintervals ($\Delta x=2$). Use the right endpoint of each strip:

Since $f$ is increasing, this right sum is an overestimate; the left sum $2(3+5+8)=32$ would be an underestimate.

As the subinterval widths shrink to zero, the Riemann sum approaches an exact value – the definite integral 定积分:

Here $\Delta x_i$ is the width of the $i$th subinterval and $x_i^{*}$ a point inside it. So a definite integral is the limit of a Riemann sum, and you should be able to translate each into the other.

A definite integral with a variable upper limit defines a new accumulation function 累积函数. The Fundamental Theorem of Calculus 微积分基本定理 (first part) says differentiation undoes this accumulation: if $f$ is continuous, then

So if $g(x)=\int_a^x f(t)\,dt$, then $g'(x)=f(x)$ and $g''(x)=f'(x)$. This is the engine behind the very common "let $g(x)=\int_a^x f(t)\,dt$" questions.

The accumulation function adds signed area; the FTC says its derivative is f

Because $g'(x)=f(x)$, everything from Unit 5 applies to an accumulation function using the graph of $f$:

• $g$ is increasing where $f>0$ and decreasing where $f<0$;
• $g$ has a local extremum where $f$ crosses zero (with a sign change);
• $g$ is concave up where $f$ is increasing; inflection points of $g$ occur where $f$ has a local extremum.

To get a value of $g$, compute the signed area: $g(x)=\int_a^x f(t)\,dt$, adding areas above the axis and subtracting areas below.

These properties simplify computation and appear constantly:

The last (splitting at an interior point $c$) lets you build a total integral from pieces read off a graph.

The second part of the Fundamental Theorem evaluates a definite integral using an antiderivative 原函数. If $F'=f$, then

So: find any antiderivative $F$, then subtract its values at the two limits. This is how most exact integrals are computed. It also gives the net change view: $\int_a^b g'(t)\,dt = g(b)-g(a)$, so a starting value plus accumulated change gives a later value, e.g. $g(5)=g(0)+\int_0^5 g'(t)\,dt$.

Worked example. Evaluate $\int_1^3 (2x+1)\,dx$. An antiderivative is $F(x)=x^2+x$, so

A definite integral is the signed area between the curve and the x-axis

An indefinite integral 不定积分 is the family of all antiderivatives, written with a constant of integration 积分常数:

Reverse each derivative rule to build the basic antiderivatives:

$u$-substitution 换元积分法 reverses the chain rule. Choose an inside function $u=g(x)$, so $du=g'(x)\,dx$, and rewrite the integral entirely in $u$:

Look for a function and its derivative both present. For a definite integral, either change the limits to $u$-values or convert back to $x$ before substituting the original limits.

Worked example. Evaluate $\int 2x\cos(x^2)\,dx$. The inside function is $u=x^2$, whose derivative $2x\,dx=du$ is present, so

Spotting that $2x$ is exactly $\dfrac{du}{dx}$ is the whole trick.

Two algebraic set-up moves let more integrals fit the basic forms: polynomial long division 多项式长除法 when the top degree is $\ge$ the bottom degree of a rational function, and completing the square 配方法 to turn a quadratic denominator into a form that integrates to an arctangent or logarithm.

A skill topic: match the integral to a method. Try a basic antiderivative first; look for a $u$-substitution (an inside function whose derivative is also present); use algebra (division, completing the square, splitting a fraction) to reshape the integrand into a standard form. Naming the structure first prevents wasted effort.

• Integration is antidifferentiation; use the power rule $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}+C$ and don't forget the $+C$.
• The Fundamental Theorem links the two: $\int_a^b f'(x)\,dx=f(b)-f(a)$, and $\tfrac{d}{dx}\int_a^x f(t)\,dt=f(x)$.
• Approximate a definite integral with Riemann sums or the trapezoidal rule from a table of values.
• A definite integral is a signed area (below the axis counts negative); split at sign changes for total area.
• Use u-substitution and remember to change the limits (or back-substitute) accordingly.