🏆Integrals

An antiderivative of f is a function whose derivative is f. Since the derivative of x³ is 3x², an antiderivative of 3x² is x³.

There is a subtlety. The derivative of x³ + 7 is also 3x², and so is the derivative of x³ - 100. Constants vanish when differentiated, so any constant can be added back. That is why an indefinite integral is written with + C:

integral of 3x² dx = x³ + C

Geometrically, all these antiderivatives are the same curve slid up and down. They have identical slopes everywhere, which is exactly what "same derivative" means.

Because the derivative rules run backwards, every antiderivative can be checked by differentiating it. That check costs seconds and catches almost every error, so it is worth doing every time.

The one antiderivative that fits. Those slid copies are a whole family of curves, one for every value of C, and a single known value of the function chooses between them. If f'(x) = 6x and f(1) = 10, then f(x) = 3x² + C for some C, and the given point turns that into one equation:

3(1)² + C = 10, so C = 7, and f(x) = 3x² + 7

Two habits make this reliable. Integrate first, substitute second: the given value is the height of the finished function at one x, not the constant itself. The two agree only when the point sits at x = 0, which is exactly why f(0) = 4 feels so much easier than f(1) = 4 — and why it hides the difference. And the constant is part of the answer here. Only one function has this derivative and passes through this point, so writing the antiderivative without its constant names a different function.

Two integrations need two conditions. When the second derivative is given, integrate twice and settle each constant the moment it appears:

f''(x) = 6x, with f'(0) = 4 and f(0) = 1 f'(x) = 3x² + C1, and f'(0) = 4 makes C1 = 4, so f'(x) = 3x² + 4 f(x) = x³ + 4x + C2, and f(0) = 1 makes C2 = 1, so f(x) = x³ + 4x + 1

The condition on f' has to be spent at the halfway stage; carrying both to the end leaves the first one with nothing to attach itself to. Motion has the same shape: a constant acceleration integrates to a velocity whose constant is the starting velocity, and again to a position whose constant is the starting position.

Differentiating multiplies by the exponent and lowers it. Integrating does the opposite: raise the exponent by one, then divide by the new exponent.

integral of xⁿ dx = xⁿ⁺¹/(n + 1) + C

integral of x⁴ dx = x⁵/5 + C integral of 6x² dx = 6 x x³/3 = 2x³ + C integral of 5 dx = 5x + C (since 5 = 5x⁰)

Sums are handled term by term, and constant multipliers stay put — just as with derivatives.

Dividing by the new exponent is the step that gets skipped. Differentiate x⁵ without it and you get 5x⁴, not x⁴: the extra factor of 5 is exactly what the division removes.

The rule has one exception. For n = -1 the new exponent would be 0 and the division would be by zero. That gap is filled by the logarithm:

integral of 1/x dx = ln(x) + C (for x > 0)

Other useful ones, all verifiable by differentiating:

integral of e^ax dx = e^ax/a + C integral of cos(ax) dx = sin(ax)/a + C integral of sin(ax) dx = -cos(ax)/a + C

A definite integral has limits and produces a number, not a function. The Fundamental Theorem of Calculus says: find any antiderivative F, then

integral from a to b of f(x) dx = F(b) - F(a)

The + C cancels in the subtraction, which is why it never appears in a definite integral.

integral from 1 to 3 of 2x dx F(x) = x² F(3) - F(1) = 9 - 1 = 8

Keep the order: top minus bottom. Reversing it flips the sign.

When f is positive on the interval, that number is the area between the curve and the x-axis. The check above is worth doing geometrically: the region under y = 2x from 1 to 3 is a trapezoid with parallel sides 2 and 6 and width 2, so its area is ((2 + 6)/2) x 2 = 8. The two methods agree, as they must.

Where the curve dips below the axis the integral counts that part as negative, so "the value of the integral" and "the total area" can differ.

The area under a straight line can be checked with a trapezoid. A curve has no such shortcut, and that is exactly where integration earns its keep.

For a function that stays above the x-axis on the interval, the area of the region between the curve and the axis is the definite integral:

area from a to b = F(b) - F(a)

Nothing new is happening — it is the same Fundamental Theorem — but two habits pay off. First, check the sign of f on the interval before calling the answer an area: where a curve dips below the axis the integral subtracts, and the two ideas part company. Second, the lower limit is not optional. When the region starts at x = 0 the term F(0) is usually 0 and can be skipped without noticing; when it starts at x = 2 it cannot.

Roots only need rewriting first, and a perfect square at the right-hand end keeps the answer exact:

integral from 0 to 9 of sqrt(x) dx = (2/3)x^3/2 evaluated from 0 to 9

Since sqrt(9) = 3, the power 9^3/2 is 3³ = 27, so the area is (2/3)(27) = 18.

For a region trapped between two graphs, integrate the gap:

area = integral from a to b of (top - bottom) dx

If a curve runs above a horizontal line y = k between the two x-values where they cross, then top minus bottom is f(x) - k, and that difference is zero at both ends — a free check that the crossings were found correctly.

y = -x² + 4x and y = 3 cross at x = 1 and x = 3

top - bottom = -x² + 4x - 3

area = integral from 1 to 3 of (-x² + 4x - 3) dx = 4/3

A rectangle of base times greatest height would have given 2 x 1 = 2 here. It overshoots, because the region is curved, and only the integral measures a curved region.

The deepest way to read an integral is as an accumulation: integrating a rate gives the total change.

If v(t) is velocity in metres per second, then the integral of v from 0 to 5 is the distance travelled in those five seconds.

v(t) = 3t² + 2 m/s s(t) = t³ + 2t distance from 0 to 5 = s(5) - s(0) = (125 + 10) - 0 = 135 m

For a constant rate you could just multiply rate by time. The integral is what handles a rate that keeps changing: it adds up rate x (tiny time) over the whole interval.

The pattern repeats across subjects. Integrate a flow rate to get a volume; integrate a rate of population growth to get the change in population; integrate acceleration to get the change in velocity. Differentiation breaks a total into its instantaneous rate; integration puts the rate back together into a total. They are the two directions of one idea.