Integration

Running the derivative backwards

Differentiating $x^3$x³ gives $3x^2$3x². Integration asks the reverse question: which function has the derivative $3x^2$3x²? One answer is $x^3$x³. A function whose derivative is f(x) is called an integral, or antiderivative, of f(x).

It is written with the integral sign: $\int 3x^{2}dx=x^3+c$∫ 3x² dx = x³ + c. The dx says that x is the variable, and the + c is explained further down. Every integral can be checked the same way: differentiate the answer and see whether the original comes back. Here the derivative of $x^3+c$x³ + c is $3x^2$3x², so it does.

Differentiating $x^3$x³: the power 3 comes down in front, then the power drops to 2, giving $3x^2$3x². Integration undoes these two moves.

Add one, then divide

Differentiating does two things in order: multiply by the power, then take one off the power. To undo them, undo the second move first. Add one to the power, then divide by the new power. In symbols, $\int x^{n}dx=\frac{x^{n+1}}{\left(n+1\right)}+c$∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c.

For $3x^2$3x², the power 2 goes up to 3, and dividing by 3 gives $\frac{3x^3}{3}=x^3$3x³/3 = x³. For $x^4$x⁴, the power goes up to 5, and dividing by 5 gives $\frac{x^5}{5}+c$x⁵/5 + c. Check by differentiating: $\frac{x^5}{5}$x⁵/5 gives $\frac{5x^4}{5}=x^4$5x⁴/5 = x⁴.

Check $\frac{x^5}{5}$x⁵/5 with a chord as well. At x = 1 its gradient should be $1^4=1$1⁴ = 1. From x = 1 to x = 1.001 the function rises from 0.2 to $1.001^5/5=0.2010020\dots$1.001⁵ / 5 = 0.2010020…, so the chord gradient is about 0.0010020 ÷ 0.001 = 1.002.

Integrating $x^4$x⁴: the power goes up from 4 to 5, then the whole term is divided by the new power 5, giving $\frac{x^5}{5}+c$x⁵/5 + c.

Numbers in front, constants and sums

A number in front stays in front, as it does when differentiating. So $\int 6x^{2}dx=6\times \frac{x^3}{3}+c=2x^3+c$∫ 6x² dx = 6 × x³/3 + c = 2x³ + c. Check: $2x^3$2x³ differentiates to $6x^2$6x².

A constant such as 6 can be written $6x^0$6x⁰, because $x^0=1$x⁰ = 1. So it integrates to $6\times \frac{x^1}{1}=6x$6 × x¹/1 = 6x, and the derivative of 6x is 6.

A sum is integrated one term at a time. For 4x + 6, the 4x gives $4\times \frac{x^2}{2}=2x^2$4 × x²/2 = 2x², and the 6 gives 6x, so $\int \left(4x+6\right)dx=2x^2+6x+c$∫ (4x + 6) dx = 2x² + 6x + c. Differentiating $2x^2+6x+c$2x² + 6x + c gives 4x + 6 back.

The rule cannot work for $x^{-1}$x⁻¹. Adding one to the power gives $x^0$x⁰, and the new power is 0, which cannot be divided by. The integral of $\frac{1}{x}$1/x is a different function, the natural logarithm, met in a later lesson.

The constant of integration

The functions $x^3$x³, $x^3+5$x³ + 5 and $x^3-2$x³ − 2 all differentiate to $3x^2$3x², because a constant differentiates to 0. So when $3x^2$3x² is integrated, there is no way to tell which constant was there. The answer is written $x^3+c$x³ + c, where c stands for any constant.

On a graph, adding c moves the whole curve up or down by c, and moving a curve straight up does not change how steep it is at any x. Take $y=x^2$y = x² and $y=x^2+3$y = x² + 3, both antiderivatives of 2x. At x = 1 the first has height 1 and the second has height 4, but both have gradient 2. Their tangents there, y = 2x − 1 and y = 2x + 2, are parallel.

The curves $y=x^2$y = x² and $y=x^2+3$y = x² + 3, one 3 units above the other. At x = 1 their tangents, y = 2x − 1 through (1, 1) and y = 2x + 2 through (1, 4), are parallel, with gradient 2.

The family $y=x^2+C$y = x² + C, with the curve for C = 2 picked out. Drag C to move the curve up or down, and drag x along it: the gradient at x stays 2x, whatever C is.

One point fixes c

If one point on the curve is known, c can be found. Suppose $\frac{dy}{dx}=2x$dy/dx = 2x and the curve passes through (1, 3). Integrating gives $y=x^2+c$y = x² + c. Put in x = 1 and y = 3: 3 = 1 + c, so c = 2 and the curve is $y=x^2+2$y = x² + 2. Check: its derivative is 2x, and at x = 1 its height is 3.

The usual mistakes

Differentiating instead of integrating. The integral of $x^4$x⁴ is not $4x^3$4x³: integration makes the power go up, to $\frac{x^5}{5}$x⁵/5.

Raising the power but not dividing. $x^5$x⁵ is not the integral of $x^4$x⁴, because its derivative is $5x^4$5x⁴, five times too big. Divide by the new power: $\frac{x^5}{5}$x⁵/5.

Dividing by the old power. $\frac{x^5}{4}$x⁵/4 differentiates to $\frac{5x^4}{4}$5x⁴/4, not $x^4$x⁴. The divisor is the new power, 5.

Leaving out the + c. Without it, the answer names one curve of the family and leaves out all the others.

Water into a tank

In the application below, water flows into a tank at a rate of 4x + 6 liters per minute. Integrating the rate gives the volume, and the water already in the tank fixes the constant.