Area as a limit of sums, the definite integral and the antiderivative

Until now you have worked with derivatives: you asked how fast a function changes. This week you ask the reverse question: if the rate of change is known, how much is the total change? The geometric counterpart is the area under a curve. You approximate the area with rectangles (a Riemann sum), send the number of rectangles to infinity (a limit), and get the definite integral. The second half of the week is the antiderivative: finding the function whose derivative is given. Next week the Fundamental Theorem of Calculus will connect the two. For this week:

1. Memorize the sigma formulas. $\sum i$, $\sum i^2$, $\sum i^3$.
2. Riemann sum: $\Delta x=\tfrac{b-a}{n}$, choose the endpoints correctly, add up the $f(x_i)\Delta x$.
3. The definite integral is signed area. Below the axis counts as negative.
4. The antiderivative table is the derivative table read backwards.
5. In an indefinite integral, $+C$ is required.

Sigma notation

$\sum_{i=1}^{n}a_i=a_1+a_2+\cdots+a_n$

Rules: $\sum(a_i\pm b_i)=\sum a_i\pm\sum b_i$, $\sum ca_i=c\sum a_i$, $\sum_{i=1}^nc=nc$. The sum of a product does not split: $\sum a_ib_i\ne\left(\sum a_i\right)\left(\sum b_i\right)$.

$\sum_{i=1}^ni=\frac{n(n+1)}{2},\qquad \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{i=1}^ni^3=\left[\frac{n(n+1)}{2}\right]^2$

Example 1. $\displaystyle\sum_{i=1}^{10}(2i^2-i)=2\cdot\frac{10\cdot11\cdot21}{6}-\frac{10\cdot11}{2}=2\cdot385-55=715$.

Riemann sums

Divide the interval $[a,b]$ into $n$ equal pieces: the width of each piece is $\Delta x=\dfrac{b-a}{n}$. Put a rectangle on each piece; its height is the value of $f$ at a point chosen inside the piece. The sum of the rectangle areas:

$\sum_{i=1}^nf(x_i^*)\,\Delta x$

Depending on where you choose the point, there are three types:

Example 2. $f(x)=4-x^2$, $[0,2]$, $n=4$. $\Delta x=0.5$.

x:      0      0.5     1      1.5     2
f(x):   4      3.75    3      1.75    0
L_4 = 0.5 (4 + 3.75 + 3 + 1.75) = 0.5 · 12.5 = 6.25
R_4 = 0.5 (3.75 + 3 + 1.75 + 0) = 0.5 · 8.5  = 4.25
true area = 16/3 ≈ 5.333

$f$ is decreasing on this interval: the left endpoint takes the highest value in each rectangle, so $L_4$ overestimates. The right endpoint underestimates. The true value is between the two.

Area = limit of sums

As $n\to\infty$ the rectangles get thinner and the sum approaches the true area. To compute it, write $x_i$ in terms of $i$, apply the sigma formulas, and take the limit.

Example 3. The area under $y=x^2$ on the interval $[0,1]$.

$\Delta x=\dfrac1n$, $x_i=\dfrac in$.

$R_n=\sum_{i=1}^n\left(\frac in\right)^2\frac1n=\frac{1}{n^3}\sum_{i=1}^ni^2=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6n^2}$

$\lim_{n\to\infty}\frac{2n^2+3n+1}{6n^2}=\frac26=\frac13$

This method is long, but it is mandatory in a question that says “use the definition / limit of Riemann sums”. With next week’s theorem the same result comes out in one line.

The definite integral

$\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\,\Delta x$

$f$ is the integrand, $a$ and $b$ are the limits. For continuous functions this limit always exists and does not depend on the choice of points.

Meaning: signed area. If $f\ge0$, it is the area under the curve. If $f$ goes below the axis, that part counts as negative.

$\int_a^bf(x)\,dx=(\text{area above the axis})-(\text{area below the axis})$

Example 4. $\displaystyle\int_{-1}^{3}(2x-2)\,dx$. The line is zero at $x=1$. On $[-1,1]$, below the axis, there is a triangle with base $2$ and depth $4$: $-4$. On $[1,3]$, above, a triangle of the same size: $+4$. The integral is $0$. But the shaded total area is $8$.

PITFALL · thinking integral = area. In a “find the area” question, if the function goes below the axis, take the absolute value of that part: $\int|f|$. If it says “evaluate the integral”, give the signed value.

Properties

$\int_a^af\,dx=0,\qquad \int_b^af\,dx=-\int_a^bf\,dx,\qquad \int_a^bf\,dx+\int_b^cf\,dx=\int_a^cf\,dx$

$\int_a^b\big(cf+g\big)\,dx=c\int_a^bf\,dx+\int_a^bg\,dx,\qquad \int_a^bc\,dx=c(b-a)$

Comparison: if $f\le g$ on $[a,b]$, then $\int_a^bf\le\int_a^bg$. If $m\le f\le M$, then $m(b-a)\le\int_a^bf\le M(b-a)$.

Integrals that can be computed geometrically: lines (triangles, trapezoids), $\sqrt{r^2-x^2}$ (a semicircle), $|x-c|$ (triangles).

The antiderivative

If $F'(x)=f(x)$, then $F$ is an antiderivative of $f$. By the consequence of the MVT (week 9), all antiderivatives of $f$ have the form $F(x)+C$. We write this with the indefinite integral:

$\int f(x)\,dx=F(x)+C$

Example 5. $\displaystyle\int(3x^2-4x+5)\,dx=x^3-2x^2+5x+C$.

Example 6 (expand first). $\displaystyle\int\frac{(x^2+1)^2}{x^2}\,dx=\int\frac{x^4+2x^2+1}{x^2}\,dx=\int\left(x^2+2+x^{-2}\right)dx=\frac{x^3}{3}+2x-\frac1x+C$.

There is no rule for the integral of a product or a quotient. First expand or divide, then integrate term by term.

TIP · check your answer by differentiating. After finding an antiderivative, take its derivative; you should get the integrand back. This 20-second check catches most sign errors on an exam.

Initial value problems

If the derivative and a point are given, $C$ is determined.

Example 7. An object has acceleration $a(t)=6t$, $v(0)=-3$, $s(0)=2$.

v(t) = ∫ 6t dt = 3t^2 + C1,      v(0) = C1 = -3     ->  v = 3t^2 - 3
s(t) = ∫ (3t^2 - 3) dt = t^3 - 3t + C2,   s(0) = C2 = 2
s(t) = t^3 - 3t + 2

Common mistakes