Running the Reel Backward Integrals

Every derivative rule learned so far answers the question, given a function, what is its rate of change. Integration asks the opposite question: given a rate of change, what was the original function. If a car's velocity at time t is known to be v(t) = 6t + 2 metres per second, what is its position function s(t)? Since differentiating s(t) = 3t squared + 2t + 5 gives exactly 6t + 2, that s(t) is one valid answer. But so is s(t) = 3t squared + 2t + 100, or s(t) = 3t squared + 2t - 40, since differentiating a constant always gives 0, so adding any constant to a correct antiderivative produces another correct antiderivative. Without extra information (such as the car's exact starting position), there is no way to pin down which specific constant is correct, only the whole family of possibilities, 3t squared + 2t + C for any constant C. This is why an antiderivative, more usually called an indefinite integral, is always written with an arbitrary constant of integration attached: the integral of f(x) dx = F(x) + C, read as 'the integral of f(x) with respect to x equals F(x) plus C,' where F is any one function whose derivative is f(x). The elongated S symbol traces back to the same idea as a sum, since (as later sections of this chapter reveal) an integral is secretly built from adding up many small pieces.

Since integration undoes differentiation, every standard derivative from the previous two chapters can be read backward into a standard integral. Because d/dx(x to the power n+1 divided by n+1) equals x to the power n (for n not equal to -1), the integral of x to the power n dx = [x to the power (n+1)] / (n+1) + C. Because d/dx(sin x) = cos x, the integral of cos(x) dx = sin(x) + C. Because d/dx(-cos x) = sin x, the integral of sin(x) dx = -cos(x) + C. Because d/dx(tan x) = sec squared(x), the integral of sec squared(x) dx = tan(x) + C. Because d/dx(e to the x) = e to the x, the integral of e to the x dx = e to the x + C. Integration also inherits linearity directly from differentiation: the integral of [f(x) plus g(x)] dx equals the integral of f(x) dx plus the integral of g(x) dx, and constants pull straight out, the integral of k times f(x) dx = k times the integral of f(x) dx. Combine these to integrate 3x squared + 4 sin(x) - 5 term by term: the integral of 3x squared dx = 3 times (x cubed/3) = x cubed; the integral of 4 sin(x) dx = 4 times (-cos x) = -4 cos(x); the integral of -5 dx = -5x, since the integral of a constant k dx is simply kx. Adding these and attaching one single overall constant gives x cubed - 4 cos(x) - 5x + C.

Many integrals are not in a standard form, but can be forced into one by a clever change of variable. To find the integral of 2x times cos(x squared) dx, notice that if u = x squared, then du/dx = 2x, so du = 2x dx exactly matches the 2x already sitting in the integral. Replace x squared with u and 2x dx with du: the integral becomes the integral of cos(u) du, a standard form, giving sin(u) + C. Substitute back u = x squared to return the answer to the original variable: sin(x squared) + C. The pattern to look for is a function and, sitting right alongside it, something proportional to that function's own derivative. Try the integral of x/(x squared + 1) dx. Let u = x squared + 1, so du = 2x dx, meaning x dx = du/2. The integral becomes the integral of (1/u) times (du/2) = (1/2) times the integral of (1/u) du = (1/2) log|u| + C, substituting back to (1/2) log|x squared + 1| + C. Substitution is often the very first thing worth trying on an unfamiliar integral, since spotting a function alongside a multiple of its own derivative is common once you know to look for it.

A rational function (one polynomial divided by another) rarely integrates directly, but splitting it into a sum of simpler fractions first often makes each piece a standard integral. To integrate 1/[(x-1)(x+2)] dx, first write 1/[(x-1)(x+2)] as A/(x-1) + B/(x+2) for constants A and B still to be found. Multiplying both sides by (x-1)(x+2) gives 1 = A(x+2) + B(x-1). Setting x = 1 (chosen to eliminate B entirely): 1 = A(3), so A = 1/3. Setting x = -2 (chosen to eliminate A entirely): 1 = B(-3), so B = -1/3. So the original fraction splits as (1/3)/(x-1) - (1/3)/(x+2), and each piece integrates directly using the standard result the integral of 1/(x-a) dx = log|x-a| + C: the integral of 1/[(x-1)(x+2)] dx = (1/3) log|x-1| - (1/3) log|x+2| + C. Partial fractions turns one intimidating fraction into a short sum of the easiest possible fractions to integrate, at the cost of a little algebra up front to find the constants.

Substitution and partial fractions do not handle every integral, particularly a product of two unrelated types of function, like x times e to the x. Integration by parts, derived directly from the product rule for derivatives run backward, handles exactly this case: the integral of u times (dv/dx) dx = u times v, minus the integral of v times (du/dx) dx, usually abbreviated to the integral of u dv = uv minus the integral of v du. The skill lies entirely in choosing which factor is u and which is dv wisely; a common guide is to let u be whichever factor gets simpler when differentiated (a polynomial like x, whose derivative is just 1), and dv be whichever factor is easy to integrate (like e to the x, which integrates to itself). Apply this to the integral of x times e to the x dx: let u = x (so du = dx) and dv = e to the x dx (so v = e to the x). The formula gives x times e to the x, minus the integral of e to the x dx = x e to the x - e to the x + C, which factors neatly as e to the x(x - 1) + C. A useful shortcut worth recognising by sight: the integral of e to the x times [f(x) + f prime(x)] dx = e to the x times f(x) + C, since the product and sum happen to combine perfectly whenever a function and its own derivative appear together multiplied by e to the x.

Every technique so far has built an indefinite integral, a family of functions. A definite integral is a specific number, the area trapped between a curve y = f(x), the x-axis, and two vertical bounds x = a and x = b, written the integral from a to b of f(x) dx. For a curve, unlike a rectangle or triangle, there is no ready-made area formula, so the area is built up directly. Slice the interval from a to b into n thin strips, each of width delta-x = (b-a)/n. Over each thin strip, approximate the curve's true, slightly-varying height with the single height value of f at that strip's left edge, turning each strip into a simple rectangle of area f(x-value) times delta-x. Adding up all n of these thin rectangles gives an approximate total area, close to but not exactly equal to the true area under the curve, since each rectangle either slightly overshoots or undershoots the actual curve within its strip. As the number of strips n is allowed to grow larger and larger, each strip becomes thinner and thinner, and the approximation using flat-topped rectangles gets closer and closer to the true, curved area, exactly the same squeezing idea used to define a derivative back in the very first chapter of this thread. The definite integral is formally defined as the limit of this sum of rectangle areas, as the number of strips n approaches infinity: the integral from a to b of f(x) dx = the limit as n to infinity of the sum of f(x_i) times delta-x, over all n strips. This is why the integral sign is an elongated S: it is literally standing in for a sum, the limiting case of adding up infinitely many, infinitely thin rectangles.

The Riemann sum defines what a definite integral means, but actually computing one by summing infinitely many rectangles directly would be painfully slow. The Fundamental Theorem of Calculus is the shortcut, and it is worth seeing why it works, not just using it. Define an area function A(x) as the running area under the curve f, starting from a fixed point a and swept out up to a variable point x. As x increases by a very small amount h, the extra sliver of area added, A(x+h) - A(x), is very close to a thin rectangle of width h and height f(x) (since over a tiny enough width, the curve barely changes height at all), so [A(x+h) - A(x)]/h is very close to f(x). As h shrinks to 0, this ratio becomes exactly A prime(x), and the approximation becomes exact: A prime(x) = f(x). In words, the derivative of the running-area function is the original function itself. But that means A(x) is an antiderivative of f(x), the very same kind of object built by every integration technique earlier in this chapter. Since any two antiderivatives of the same function differ only by a constant, A(x) = F(x) + C for whatever antiderivative F you already know how to find algebraically. Because A(a) = 0 (no area has been swept out yet at the very start), the constant works out to C = -F(a), giving A(x) = F(x) - F(a). Setting x = b gives the running area all the way from a to b: A(b) = F(b) - F(a). This is the entire content of the Fundamental Theorem of Calculus: the integral from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f. An infinite sum of shrinking rectangles turns out to be exactly computable by finding one antiderivative and subtracting its values at the two endpoints, because the running area's own rate of growth was f(x) all along.

In practice, evaluating a definite integral almost never needs a Riemann sum at all. Find any antiderivative F(x), then compute F(b) - F(a), a calculation usually written F(x) evaluated between a and b. Evaluate the integral from 1 to 3 of (2x + 1) dx: an antiderivative is x squared + x, so the definite integral is (3 squared + 3) - (1 squared + 1) = 12 - 2 = 10. A handful of properties of definite integrals make many calculations shorter. Swapping the limits flips the sign: the integral from b to a of f(x) dx = negative of the integral from a to b of f(x) dx. An integral can be split at any interior point c between a and b: the integral from a to b = the integral from a to c, plus the integral from c to b. And a genuinely useful symmetry: if f is an even function (f(-x) = f(x), symmetric about the y-axis), the integral from -a to a of f(x) dx = 2 times the integral from 0 to a of f(x) dx, since the two symmetric halves contribute equally; but if f is odd (f(-x) = -f(x), symmetric through the origin), the integral from -a to a of f(x) dx = 0 outright, since the negative half exactly cancels the positive half. Evaluate the integral from -2 to 2 of x cubed dx without any algebra at all: x cubed is an odd function (since (-x) cubed = -(x cubed)), so the integral is 0 by the symmetry property alone, no antiderivative needed.