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Definite Integrals



The Riemann Sum Construction
Definite Integral Notation
Signed Area Interpretation
Properties of Definite Integrals
Linearity
Computing Definite Integrals
Average Value of a Function
Summary: Three Lenses on the Definite Integral
Common Pitfalls
Definite Integrals FAQ



From Sums to Areas


The definite integral answers a concrete question: what is the total accumulated quantity between two points? Geometrically, this corresponds to the area between a curve and the horizontal axis—but with a crucial refinement. Regions below the axis contribute negative area, making the integral a signed quantity.

The notation

Summary: Three Lenses on the Definite Integral
∫abf(x) dx\int_a^b f(x)\, dx

specifies the integrand f(x)f(x), the variable of integration xx, and the bounds from aa to bb. The result is a single number, not a function. This number represents net accumulation: the sum of infinitely many infinitesimal contributions f(x) dxf(x)\, dx as xx traverses the interval.

Riemann sums provide the rigorous foundation. Approximate the region with rectangles, compute their total area, and take the limit as the rectangles become infinitely thin. What emerges is not an approximation but an exact value—the definite integral.

Key Terms

Definite Integral— ∫abf(x) dx\int_a^b f(x)\,dx, a number representing accumulated quantity
Riemann Sum— the approximating sum whose limit is the integral
Bounds of Integration— where accumulation starts (aa) and ends (bb)
Signed Area— positive above the axis, negative below
Average Value of a Function— 1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx
Integrand— the function being accumulated

See All Calculus Definitions →


The Riemann Sum Construction


The definite integral arises as a limit of approximating sums.

Partition the interval [a,b][a, b] into nn subintervals of width Δx=(b−a)/n\Delta x = (b - a)/n. In each subinterval, choose a sample point xi∗x_i^* and form the rectangle with height f(xi∗)f(x_i^*) and width Δx\Delta x. The total area of these rectangles is the Riemann sum:

The Riemann Sum Construction
Sn=∑i=1nf(xi∗)ΔxS_n = \sum_{i=1}^{n} f(x_i^*) \Delta x

As n→∞n \to \infty and the rectangles become infinitely thin, the Riemann sum approaches the definite integral:

The Riemann Sum Construction
∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗)Δx\int_a^b f(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x

THEOREM
The choice of sample points—left endpoints, right endpoints, midpoints—affects individual Riemann sums but not the limit, provided ff is integrable.

Fact: Archimedes summed a parabola first
In the third century BC Archimedes proved that a parabolic segment has 43\frac{4}{3} the area of its inscribed triangle. He filled it with triangles whose areas form 1+14+116+⋯1 + \frac{1}{4} + \frac{1}{16} + \cdots times the first one, and he showed that this sum is 43\frac{4}{3} exactly.

Filling a curved region with simpler shapes of known area is far older than the integral sign. Greek geometers did it by what is now called the method of exhaustion: pieces are added until the region left over is smaller than any given amount, and a proposed value is confirmed by showing that the area can be neither larger nor smaller than it. A Riemann sum does the same with thin rectangles, and the limit replaces the two-sided argument with a single step.
Sample point choice Formula for xi* Resulting Riemann sum
Left endpoint xi* = xi−1  (left edge of i-th subinterval) left Riemann sum — rectangle heights f(xi−1)
Right endpoint xi* = xi  (right edge of i-th subinterval) right Riemann sum — rectangle heights f(xi)
Midpoint xi* = (xi−1 + xi) / 2 midpoint Riemann sum — rectangle heights at the center of each subinterval
1 2 3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 x y f(x) = x²
Left-endpoint sum for x² on [0, 3], n = 8

The interval is cut into eight strips of width Δx = 3/8 and each rectangle takes the height of the curve at the left end of its strip; because the parabola rises, every rectangle falls short and the sum underestimates the area, 9. Choose the right endpoints instead and every rectangle overshoots. Switch the sample point and increase n on the Riemann sum visualizer.

The definite integral is the value all of these sums share once n grows without bound.

Definite Integral Notation

Notation

Definite Integral Notation

What the bounds add to the sign, which letters are disposable, and the bracket that closes every FTC computation.
The bare sign and dxdx — indefinite integrals; Δx\Delta x — differentials; lim⁡\lim — two-sided limits.
∫abf(x) dx\int_a^b f(x)\, dx
The integral from a to b of f of x, d x
The indefinite sign with bounds welded on — and the output changes species: a number, not a family. Lower limit aa below, upper limit bb above; sign and bounds act as one operator on the integrand.
Casesb<ab < a is legal — reversing the bounds flips the sign, Properties of Definite Integrals below. a=ba = b gives 00. A bound of ±∞\pm\infty, or an integrand blowing up inside [a,b][a,b], hands the symbol to improper integrals.
Also written∫[a,b]f\int_{[a,b]} f — over-the-set style in analysis and measure theory, where orientation is deliberately absent.
Do not confuseThe indefinite family. No +C+C appears here — any antiderivative gives the same F(b)−F(a)F(b) - F(a); the constant cancels.
∫01t2 dt=∫01x2 dx\int_0^1 t^2\, dt = \int_0^1 x^2\, dx
The variable of integration is a dummy
Inside a definite integral the letter is a bound placeholder — it vanishes with the dxdx when the number comes out. Both spellings above name the same value, 13\tfrac{1}{3}; so would uu, τ\tau, or anything else.
CasesThe letter turns load-bearing in accumulation functions: ∫axf(t) dt\int_a^x f(t)\,dt needs tt inside because xx is busy in the bound — ∫axf(x) dx\int_a^x f(x)\,dx collides the two roles. Engineering writes τ\tau under the sign when tt is real time outside it.
Also writtenξ\xi as the standing dummy in Russian analysis texts.
Do not confuseThe indefinite integral's variable — not a dummy there: the answer F(x)+CF(x) + C keeps it alive.
F(x) ∣abF(x)\,\Big|_a^b · [F(x)]ab\left[F(x)\right]_a^b
F of x, evaluated from a to b
Shorthand for F(b)−F(a)F(b) - F(a) — the closing move of every FTC computation, Computing Definite Integrals below. Upper value minus lower, always that order.
CasesBar and bracket forms are interchangeable; the bracket survives long expressions better in print. Reversing the bounds flips its sign along with the integral's.
Also writtenF∣abF\big|_a^b with the argument suppressed.
Do not confuseThe one-point evaluation bar ∣x=a\big|_{x=a} — same glyph, different arithmetic: substitution there, subtraction of two substitutions here.
∑i=1nf(xi∗) Δx  ⟶  ∫abf(x) dx\sum_{i=1}^{n} f(x_i^*)\,\Delta x \;\longrightarrow\; \int_a^b f(x)\, dx
The Riemann sum becomes the integral
The notation records its own limit, from The Riemann Sum Construction above: Σ\Sigma stretches into the elongated S, Δx\Delta x shrinks to dxdx, the sampled xi∗x_i^* relaxes into a continuous xx. Leibniz designed the symbol as exactly this metamorphosis.
CasesThe star in xi∗x_i^* marks a chosen sample point — left endpoint, right endpoint, midpoint all legal; the limit forgets the choice.
Also writtenSnS_n for the nn-th Riemann sum, as above.
Same glyph elsewhereThe same superscript star marks the complex conjugate z∗z^* — an unrelated job in the same corner of the symbol.
-5 -4 -3 -2 -1 1 2 3 4 5 -1 1 2 x y f(t) = sin(t) F(x) = −cos(x) + cos(a)
sin t from 0 to 2: area counted with sign

The shading follows the curve above the axis, and if the right edge were pushed past π the shading below the axis would count as negative: the accumulated value rises while the curve is above the axis and falls while it is below. Here, from 0 to 2, the total is 1 − cos 2 ≈ 1.42. Push the right edge past π and watch the total start to shrink on the fundamental theorem of calculus visualizer.

Total unsigned area needs the integrand split at its zeros; the integral itself does not.

Signed Area Interpretation


The definite integral computes area with sign.

Where f(x)>0f(x) > 0, the region between the curve and the xx-axis lies above the axis and contributes positive area. Where f(x)<0f(x) < 0, the region lies below the axis and contributes negative area.

The integral sums these signed contributions:

Signed Area Interpretation
∫abf(x) dx=(area above)−(area below)\int_a^b f(x)\, dx = (\text{area above}) - (\text{area below})

This means the integral can be zero even when substantial area exists—positive and negative regions may cancel. It can also be negative when the curve lies predominantly below the axis.

To find total unsigned area, integrate the absolute value:

Total Unsigned Area
Total area=∫ab∣f(x)∣ dx\text{Total area} = \int_a^b |f(x)|\, dx
Learn more about this formula: Total Unsigned Area →


Pitfall: Moving the absolute value outside
Taking ∣∫−12x dx∣=32\left|\int_{-1}^{2} x\, dx\right| = \frac{3}{2} as the area still lets the part below the axis cancel against the part above. Splitting at the zero x=0x = 0 gives 12+2=52\frac{1}{2} + 2 = \frac{5}{2}.

Computing ∫ab∣f(x)∣ dx\int_a^b |f(x)|\, dx starts with the zeros of ff in [a,b][a, b]. Between consecutive zeros ff keeps one sign, so on each piece ∣f(x)∣|f(x)| is either f(x)f(x) or −f(x)-f(x) and an ordinary antiderivative applies. The integrals over the pieces are then added.
sin x on [0, 2 pi]: area +2 above the axis and -2 below cancel, so the integral is 0xy1234567−11y = sin x+2−2integral over [0, 2π]: 2 − 2 = 0Signed area: the part below the axis counts negative
Area above and below the axis

y = sin x on [0, 2π]. The blue strips between the curve and the axis on [0, π] have total area 2 and count as +2. The amber strips on [π, 2π] have the same area but lie below the axis, so they count as −2. The definite integral adds the two: 2 − 2 = 0, although the total shaded area is 4, which is what ∫|sin x| dx would give.

Over [0, π] alone the integral is 2: only a full period cancels.
A definite integral measures net accumulation, not area: to count every region as positive, integrate ∣f∣|f|.

Properties of Definite Integrals


Definite integrals satisfy several fundamental properties.

Additivity over intervals:

Additivity Over Intervals
∫abf(x) dx+∫bcf(x) dx=∫acf(x) dx\int_a^b f(x)\, dx + \int_b^c f(x)\, dx = \int_a^c f(x)\, dx
Learn more about this formula: Additivity Over Intervals →


Reversing limits negates the integral:

Reversing Limits of Integration
∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx
Learn more about this formula: Reversing Limits of Integration →


Pitfall: Ignoring the order of the limits
Treating ∫312x dx\int_3^1 2x\, dx as the integral over [1,3][1, 3] gives 9−1=89 - 1 = 8. The upper limit is 11, so the value is F(1)−F(3)=1−9=−8F(1) - F(3) = 1 - 9 = -8.

Integrating from bb down to aa makes the width Δx=(a−b)/n\Delta x = (a - b)/n negative, so every term f(xi∗) Δxf(x_i^*)\,\Delta x of the Riemann sum changes sign while the heights stay the same. The bounds therefore carry a direction as well as an interval: the lower limit is where accumulation starts and the upper limit is where it stops, even when the upper limit is the smaller number.

Zero-width interval:

Zero-Width Interval
∫aaf(x) dx=0\int_a^a f(x)\, dx = 0
Learn more about this formula: Zero-Width Interval →


Comparison: If f(x)≤g(x)f(x) \leq g(x) on [a,b][a, b], then

Properties of Definite Integrals
∫abf(x) dx≤∫abg(x) dx\int_a^b f(x)\, dx \leq \int_a^b g(x)\, dx

Linearity


Definite integrals respect addition and scalar multiplication.

Sum rule:

Linearity
∫ab[f(x)+g(x)] dx=∫abf(x) dx+∫abg(x) dx\int_a^b [f(x) + g(x)]\, dx = \int_a^b f(x)\, dx + \int_a^b g(x)\, dx

Constant multiple rule:

Linearity
∫abc⋅f(x) dx=c∫abf(x) dx\int_a^b c \cdot f(x)\, dx = c \int_a^b f(x)\, dx

Pitfall: Integrating a product factor by factor
Splitting ∫01x⋅x dx\int_0^1 x \cdot x\, dx into ∫01x dx⋅∫01x dx\int_0^1 x\, dx \cdot \int_0^1 x\, dx gives 12⋅12=14\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. Linearity covers sums and constant factors only; ∫01x2 dx=13\int_0^1 x^2\, dx = \frac{1}{3}.

These rules allow complex integrands to be broken into simpler pieces, each integrated separately and then combined.

Computing Definite Integrals


Direct computation via Riemann sums is tedious. The Fundamental Theorem of Calculus provides the shortcut.

If FF is any antiderivative of ff—meaning F′(x)=f(x)F'(x) = f(x)—then:

Computing Definite Integrals
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\, dx = F(b) - F(a)

Pitfall: Dropping the lower-limit term
Assuming F(0)=0F(0) = 0 in ∫0πsin⁡x dx=[−cos⁡x]0π\int_0^{\pi} \sin x\, dx = \left[-\cos x\right]_0^{\pi} gives −cos⁡π=1-\cos \pi = 1. But F(0)=−cos⁡0=−1F(0) = -\cos 0 = -1, so the integral is 1−(−1)=21 - (-1) = 2.

This result, detailed on the rules page, transforms integration from a limiting process into a two-step procedure: find an antiderivative, then evaluate at the endpoints.

The notation F(x)∣abF(x) \Big|_a^b or [F(x)]ab[F(x)]_a^b denotes the evaluation F(b)−F(a)F(b) - F(a).
-5 5 -20 -15 -10 -5 5 10 15 20 x y f(t) = t² F(x) = x³/3 − a³/3
∫₀² t² dt = F(2) − F(0) = 8/3

The shaded area is the definite integral, and the accumulation curve above it reads F(2) = 8/3 exactly: the antiderivative t³/3 evaluated at the right edge minus its value at the left. No rectangles were summed to get the number. Move the right edge and read the area straight off the antiderivative on the fundamental theorem of calculus visualizer.

This is why antiderivatives, not rectangles, do the computing in practice.

Average Value of a Function


The average value of ff on the interval [a,b][a, b] is:

Average Value of a Function
favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\, dx
Learn more about this formula: Average Value of a Function →


Pitfall: Averaging the endpoint values
Taking f(0)+f(3)2\frac{f(0) + f(3)}{2} for f(x)=x2f(x) = x^2 on [0,3][0, 3] gives 0+92=4.5\frac{0 + 9}{2} = 4.5. The average value uses the whole interval: 13∫03x2 dx=93=3\frac{1}{3}\int_0^3 x^2\, dx = \frac{9}{3} = 3.

This generalizes the familiar average of discrete values. The integral computes the total, and division by the interval length yields the mean.

Geometrically, favgf_{\text{avg}} is the height of a rectangle with base [a,b][a, b] whose area equals the area under the curve. The Mean Value Theorem for Integrals guarantees that a continuous function actually attains this average value at some point cc in (a,b)(a, b):

Average Value of a Function
f(c)=favgf(c) = f_{\text{avg}}

Fact: Mains at 230 volts peaks near 325
An alternating voltage is rated instead by the square root of the average of its square. The average of sin⁡2x\sin^2 x over a cycle is one half, so the rating is the peak divided by 2\sqrt{2}, and 230230-volt mains swings to about ±325\pm 325 volts.

The average value depends on the interval as well as on the function. On [0,π][0, \pi] the arch of sin⁡x\sin x has area 22, so its average value there is 2π≈0.64\frac{2}{\pi} \approx 0.64. On the full period [0,2π][0, 2\pi] the area below the axis cancels the area above, and the average value is 00. Any quantity that swings evenly between positive and negative values has average value 00 over a whole cycle, however large its swings are.
x^2 on [0, 3] has average value 3: the excess above y = 3 balances the shortfall below; reached at x = sqrt 3xy1234−112345678910f_avg = 3y = x²c = √3area under x² on [0, 3]: 9 = 3 · 3amber excess = navy shortfallThe average value is the level where excess and shortfall balance
The average value as a balancing level

f(x) = x² on [0, 3] has ∫₀³ x² dx = 9, so its average value is 9/3 = 3. The dashed level y = 3 is that average: where the curve is above it (amber) the excess area exactly equals the shortfall where the curve is below it (navy), so a rectangle of height 3 over [0, 3] has the same area 9. The curve crosses the level at c = √3, the point the Mean Value Theorem for Integrals promises.

Because x² curves upward, its average 3 lies above the midpoint value f(1.5) = 2.25.
The average value is the total divided by the interval's length, and a continuous function always attains it.

Summary: Three Lenses on the Definite Integral


The sections above introduce the definite integral from three angles: as the limit of Riemann sums (the construction), as signed area between a curve and the x-axis (the geometric meaning), and as a way to define the average value of a function on an interval (a derived use). The table below collects these three lenses together — each is a valid perspective, and each is useful in a different setting. Computing a definite integral in practice usually relies on the Fundamental Theorem of Calculus, detailed on the rules page.
Lens Defining expression What it gives Geometric picture
Riemann sum (limit definition) limn → ∞ Σi=1n f(xi*) Δx the rigorous construction — what the integral fundamentally is infinitely many rectangles becoming infinitesimally thin
Signed area ∫ab f(x) dx a single number measuring net accumulation over [a, b] area above the x-axis minus area below
Average value (1 / (b − a)) · ∫ab f(x) dx the mean output of f over [a, b] height of a rectangle on [a, b] whose area equals the area under f

Common Pitfalls


Five mistakes cause many wrong answers with definite integrals. Each is marked where it happens in the sections above. Here each is worked through in full, with the wrong and the right working side by side.

Moving the absolute value outside

The total area between a curve and the axis is wanted, and the absolute value is applied to the finished integral instead of to the integrand. The cancellation between the parts above and below the axis has then already happened.

Wrong working: ∣∫−12x dx∣=∣[x22]−12∣=∣2−12∣=32\left|\int_{-1}^{2} x\, dx\right| = \left|\left[\frac{x^2}{2}\right]_{-1}^{2}\right| = \left|2 - \frac{1}{2}\right| = \frac{3}{2}.

Right working: xx is negative on (−1,0)(-1, 0) and positive on (0,2)(0, 2), so split at 00. ∫−10(−x) dx=[−x22]−10=0−(−12)=12\int_{-1}^{0} (-x)\, dx = \left[-\frac{x^2}{2}\right]_{-1}^{0} = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2} and ∫02x dx=2\int_0^2 x\, dx = 2, so the total area is 12+2=52\frac{1}{2} + 2 = \frac{5}{2}.

Why it happens: ∣∫abf(x) dx∣\left|\int_a^b f(x)\, dx\right| and ∫ab∣f(x)∣ dx\int_a^b |f(x)|\, dx look almost the same, and for a function that keeps one sign on [a,b][a, b] they are equal.

How to check: the region here is two triangles, one with base 11 and height 11, area 12\frac{1}{2}, and one with base 22 and height 22, area 22. In general ∣∫abf(x) dx∣≤∫ab∣f(x)∣ dx\left|\int_a^b f(x)\, dx\right| \leq \int_a^b |f(x)|\, dx, with equality only when ff does not change sign.

Ignoring the order of the limits

An integral whose upper limit is smaller than its lower limit is evaluated as if it ran over the same interval in the usual direction.

Wrong working: ∫312x dx\int_3^1 2x\, dx is read as an integral over [1,3][1, 3], so it equals [x2]13=9−1=8\left[x^2\right]_1^3 = 9 - 1 = 8.

Right working: the upper limit is 11 and the lower limit is 33, so ∫312x dx=F(1)−F(3)=1−9=−8\int_3^1 2x\, dx = F(1) - F(3) = 1 - 9 = -8, which is −∫132x dx-\int_1^3 2x\, dx.

Why it happens: both versions cover the same set of points, so the order of the bounds seems to be only a matter of notation.

How to check: when the integrand is positive and the upper limit is the smaller number, the integral must come out negative.

Integrating a product factor by factor

The sum rule and the constant multiple rule are extended to products, as if the integral of a product were the product of the integrals.

Wrong working: ∫01x⋅x dx=∫01x dx⋅∫01x dx=12⋅12=14\int_0^1 x \cdot x\, dx = \int_0^1 x\, dx \cdot \int_0^1 x\, dx = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.

Right working: simplify the integrand first. x⋅x=x2x \cdot x = x^2, and ∫01x2 dx=[x33]01=13\int_0^1 x^2\, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}.

Why it happens: both rules of linearity let the integral pass through a combination of functions, and a product looks like the next case of the same pattern.

How to check: try the simplest functions. With f(x)=g(x)=1f(x) = g(x) = 1 on [0,2][0, 2], ∫021⋅1 dx=2\int_0^2 1 \cdot 1\, dx = 2, while ∫021 dx⋅∫021 dx=2⋅2=4\int_0^2 1\, dx \cdot \int_0^2 1\, dx = 2 \cdot 2 = 4.

Dropping the lower-limit term

After the antiderivative is found, only its value at the upper limit is kept. The habit comes from examples where the lower limit is 00 and the antiderivative is 00 there.

Wrong working: ∫0πsin⁡x dx=[−cos⁡x]0π=−cos⁡π=1\int_0^{\pi} \sin x\, dx = \left[-\cos x\right]_0^{\pi} = -\cos \pi = 1.

Right working: ∫0πsin⁡x dx=(−cos⁡π)−(−cos⁡0)=1−(−1)=2\int_0^{\pi} \sin x\, dx = (-\cos \pi) - (-\cos 0) = 1 - (-1) = 2.

Why it happens: in ∫02t2 dt=[t33]02\int_0^2 t^2\, dt = \left[\frac{t^3}{3}\right]_0^2 the lower term is 00, and the step is skipped by habit. A lower limit of 00 does not make F(0)F(0) equal to 00.

How to check: estimate the area. The arch of sin⁡x\sin x on [0,π][0, \pi] lies above the triangle with base π\pi and height 11, whose area is π2≈1.57\frac{\pi}{2} \approx 1.57, so 11 is too small.

Averaging the endpoint values

The average value of a function on [a,b][a, b] is taken as the average of its two endpoint values, as for two numbers, instead of from the integral over the whole interval.

Wrong working: for f(x)=x2f(x) = x^2 on [0,3][0, 3], favg=f(0)+f(3)2=0+92=4.5f_{\text{avg}} = \frac{f(0) + f(3)}{2} = \frac{0 + 9}{2} = 4.5.

Right working: favg=13−0∫03x2 dx=13⋅9=3f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2\, dx = \frac{1}{3} \cdot 9 = 3.

Why it happens: the average of two numbers is familiar, and for a linear function it gives the right answer. For f(x)=2x+1f(x) = 2x + 1 on [0,3][0, 3] both methods give 44, which hides the difference.

How to check: the rectangle of height favgf_{\text{avg}} over [a,b][a, b] must have the same area as the region under the curve. 4.5⋅3=13.54.5 \cdot 3 = 13.5, but ∫03x2 dx=9\int_0^3 x^2\, dx = 9; 3⋅3=93 \cdot 3 = 9 matches.

Definite Integrals FAQ

Why is there no + C in a definite integral?

+
Because the constant cancels. A definite integral evaluates F(b) − F(a), and using F(x) + C instead gives (F(b) + C) − (F(a) + C), where the two copies of C subtract away. Any antiderivative produces the same number, so there is nothing for + C to record. The family only survives in the indefinite integral.Read more →

Why is ∫ₐˣ f(t) dt written with t instead of x?

+
Because x is already in use as the upper bound. Inside a definite integral the variable is a dummy — ∫₀¹ t² dt and ∫₀¹ x² dx name the same number — so the letter is normally free. But an accumulation function needs x for the bound, and ∫ₐˣ f(x) dx would give one letter two jobs at once.Read more →

Does it matter whether you use left or right endpoints in a Riemann sum?

+
For any fixed number of rectangles, yes — left endpoints, right endpoints, and midpoints generally give three different numbers. In the limit, no: as the rectangles become infinitely thin all three converge to the same definite integral, provided f is integrable. The star in x* marks the sample point precisely because the choice is free.Read more →

How do you find the total area between a curve and the x-axis?

+
Integrate the absolute value: total area = ∫ₐᵇ |f(x)| dx. The plain definite integral gives signed area, counting regions below the axis as negative, so it can come out smaller than the true area or even zero when the parts cancel. In practice, find where f crosses the axis, integrate each piece separately, and add the absolute values.Read more →

What is the Mean Value Theorem for Integrals?

+
It guarantees that a continuous function actually attains its own average value somewhere on the interval: there is at least one c in (a, b) with f(c) = f_avg, where f_avg = (1/(b − a))∫ₐᵇ f(x) dx. Geometrically, the rectangle of height f(c) over [a, b] has exactly the area under the curve.Read more →