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Approximating an area with rectangles and defining the definite integral as their limit; signed area and accumulation; and the two parts of the fundamental theorem.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can compute left and right Riemann sums, say which side of the truth each falls on and why, evaluate an integral by geometry, and read an integral of a rate as an accumulated quantity with the right units. You can also use both parts of the fundamental theorem: evaluating a definite integral from an antiderivative, and differentiating an accumulation function without integrating anything.
You can find the area of a rectangle, and you can take a limit. Those two are enough to define the area under any curve — which is what the definite integral is, before any technique for computing one appears.
Riemann sum: a total of (height $\times$ width) over strips approximating an area.
Left / right sum: the height read at the left or right edge of each strip.
Definite integral $\int_a^b f(x)\,dx$: the limit of those sums as the strips shrink.
Signed area: area below the axis counts as negative.
Accumulation: integrating a rate to get the quantity it produces.
The area under a curve has no formula, so it is defined by approximation. Cut $[a, b]$ into $n$ strips of width $\Delta x = \dfrac{b-a}{n}$, take a height in each — the left edge, the right edge, the middle — and add up height times width. That total is a Riemann sum, and the definite integral is its limit as $n \to \infty$:
$$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x$$
The elongated S is a stretched 'sum', and $dx$ is what $\Delta x$ becomes in the limit. Every part of the notation is a record of where the definition came from.
Two things follow immediately. First, an integral is a signed area: where $f$ is negative, so is its contribution. Second — and this is what makes integration useful outside geometry — integrating a rate accumulates the quantity. Litres per minute integrated over minutes gives litres; metres per second over seconds gives metres. The units multiply, exactly as height times width does, because that is what the sum is doing.
For an increasing function the left sum is an under-estimate and the right sum an over-estimate, so together they bracket the truth. For a decreasing one the roles swap. It is worth deriving that each time from the picture rather than memorising it, because the memorised version is wrong half the time.
Another way: picture
Picture the region under a curve filled with tall thin rectangles, each touching the curve at one edge. Some poke out above it, some fall short. Now halve their width and double their number: the total error halves too. Keep going, and the sum has nowhere to settle but the area itself.
Another way: steps
"A left sum always under-estimates." Only for an increasing function. For a decreasing one it over-estimates, and for a function that turns it may do either. Which side of the truth a sum falls on is a fact about the shape.
"An integral is an area, so it cannot be negative." It is a signed area. Where the function is below the axis the contribution is negative, and an integral over a symmetric interval can come out zero for a function that is never zero.
"Integrating a rate gives the total." It gives the change. Whatever was there at the start has to be added, and forgetting it is the single most common error in accumulation problems.
Width $= \tfrac{2}{4} = 0.5$.
Interval over strips.
Left heights at $0, 0.5, 1, 1.5$: $0, 0.25, 1, 2.25$, totalling $3.5$; times $0.5$ gives $1.75$.
Left edges.
Right heights at $0.5, 1, 1.5, 2$: total $6$; times $0.5$ gives $3$.
Right edges.
The exact value is $\tfrac{8}{3} \approx 2.67$, comfortably between them.
Bracketed, as an increasing function must be.
$\int_0^4 3x\,dx$ is the area of a triangle with base $4$ and height $12$.
Sketch first.
$\tfrac{1}{2}(4)(12) = 24$.
No calculus needed.
$\int_{-4}^{4} 3x\,dx = 0$: the triangle below the axis cancels the one above.
Signed area.
Geometry: a triangle, base $3$, height $6$, area $9$.
Right sum: width $1$, heights at $1, 2, 3$ are $2, 4, 6$, total $12$.
The sum over-estimates by $3$, as it must for an increasing function — and halving the width would roughly halve that error.
Approximate $\displaystyle\int_0^{4} x^2 \, dx$ with a left Riemann sum using $6$ equal strips.
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Approximate $\displaystyle\int_0^{3} x \, dx$ with a right Riemann sum using $2$ equal strips.
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For $\displaystyle\int_0^{4} x^3 \, dx$, the left sum with $6$ strips is $400/9$ and the right sum is $784/9$. Where is the true value?
Find $\displaystyle\int_0^{4} 4x \, dx$ by geometry.
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You can define a definite integral as a limit of sums, and you can differentiate anything in this course. The theorem in this section connects those two — and turns the computation of an area from an infinite process into a subtraction.
Antiderivative $F$: a function with $F' = f$.
Indefinite integral $\int f\,dx = F(x) + C$: the whole family of them.
Definite integral $\int_a^b f$: a number, and the $C$ cancels.
Average value: $\dfrac{1}{b-a}\int_a^b f$.
Accumulation function: $g(x) = \int_a^x f$, an integral with a variable upper limit.
The fundamental theorem of calculus has two parts, and they run in opposite directions.
Part one — evaluation. If $F$ is any antiderivative of $f$, then
$$\int_a^b f(x)\,dx = F(b) - F(a)$$
An area, defined as the limit of infinitely many rectangles, is computed by two substitutions and a subtraction. It is difficult to overstate how large a shortcut this is: before it, each area was a separate feat of ingenuity.
Part two — differentiation. If $g(x) = \displaystyle\int_a^x f(t)\,dt$, then $g'(x) = f(x)$. Differentiating an accumulation gives back the rate that was accumulating. And note what this guarantees: every continuous function has an antiderivative, namely its own accumulation function — even the ones, like $e^{-x^2}$, whose antiderivative no combination of familiar functions can express.
Two things follow that are worth having separately. Antidifferentiation reverses the power rule: $\int x^n dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \ne -1$. And the average value of $f$ on $[a, b]$ is $\dfrac{1}{b-a}\int_a^b f$ — the height of the rectangle with the same base and the same area, which the function actually attains somewhere in the interval.
Another way: picture
Picture a tap filling a tank. The rate of flow is $f$; the amount in the tank is its accumulation $g$. Part two says that how fast the tank is filling right now is exactly the current flow — obvious, put that way. Part one says the water added between two times is the tank level at the second minus the level at the first, which is equally obvious and, written as an integral, is a theorem.
Another way: steps
| Part one | Part two | |
|---|---|---|
| Statement | $\int_a^b f = F(b) - F(a)$ | $\dfrac{d}{dx}\int_a^x f = f(x)$ |
| Direction | Integral, computed from an antiderivative | Derivative of an integral |
| What it gives you | A number | A function |
| What it guarantees | Areas are computable | Every continuous function has an antiderivative |
The last cell of the last column is the deeper half. Part one is a method; part two is an existence theorem, and it covers functions no method can reach.
*"$F(b) - F(a)$ needs the right antiderivative."* Any of them works. Adding $C$ adds it at both ends and it cancels — which is exactly why the constant is dropped for a definite integral and never for an indefinite one.
"$\dfrac{d}{dx}\int_a^x f$ requires working out the integral first." It does not: the answer is $f(x)$, immediately. That is the content of the second part, and it holds even when no antiderivative can be written down.
"The average value is the average of $f(a)$ and $f(b)$." It is the integral divided by the width. Those agree for a straight line and for almost nothing else.
$\int_0^3 6x^2\,dx$: an antiderivative is $2x^3$.
Reverse the power rule.
$2(27) - 2(0) = 54$.
Top minus bottom.
Using $2x^3 + 7$ instead gives $(54 + 7) - (0 + 7) = 54$ again.
The constant always cancels.
$f(x) = 6x^2$ on $[0, 3]$; the integral is $54$.
From above.
Divide by the width $3$: the average value is $18$.
$f(x) = 18$ when $x = \sqrt{3} \approx 1.73$, which is inside the interval — as the mean value theorem for integrals promises.
An antiderivative of $4x^3$ is $x^4$.
$16 - 1 = 15$.
Top minus bottom.
And the derivative of the accumulation is just $4x^3$ — no integration required, by part two.
Evaluate $\displaystyle\int_0^{1} 18x^2 \, dx$.
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Let $g(x) = \displaystyle\int_0^{x} 5t^2 \, dt$. Find $g'(1)$.
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Find the average value of $f(x) = 18x^2$ on $[\,0, 5\,]$.
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A tank's flow rate is $7$ litres a minute for the first $2$ minutes, then $-4$ litres a minute for the next $2$ minutes. The tank started with $70$ litres. How much does it hold at the end?
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Water flows into a tank at $R(t)$ litres per minute. What does $\displaystyle\int_{9}^{15} R(t)\, dt$ represent?
Complete the two parts of the fundamental theorem: the first says $\int_a^b f = F(b) - F(a)$ where $F$ is an ___ of $f$; the second says $\dfrac{d}{dx}\int_a^x f = $ ___ .
a; b
You can approximate, interpret and evaluate definite integrals. Without looking: why does the constant of integration not matter in a definite integral, and what is $\dfrac{d}{dx}\int_a^x f(t)\,dt$?
8. Your turn: $\int_0^3 2x\,dx$ by geometry, then by a right sum with 3 strips, step 3
20. Your turn: $\int_1^2 4x^3\,dx$, and then $\dfrac{d}{dx}\int_1^x 4t^3\,dt$, step 3