Properties of the Definite Integral

Reversing the limits

The integral from a to b of f(x) dx is F(b) − F(a). Run it the other way, from b to a, and it becomes F(a) − F(b), which is −(F(b) − F(a)). So swapping the limits changes the sign.

In strips: going from b back to a, each step is $\Delta x=\left(a-b\right)/n$Δx = (a − b) / n, a negative width, so every term $f\left(x_i\right)\Delta x$f(xᵢ) Δx changes sign and so does the sum.

Check it on $y=\frac{x^2}{4}$y = x²/4, whose antiderivative is $\frac{x^3}{12}$x³/12. From 0 to 2 the integral is $\frac{8}{12}-0=\frac{2}{3}$8/12 − 0 = 2/3. From 2 to 0 it is $0-\frac{8}{12}=-\frac{2}{3}$0 − 8/12 = −2/3.

Equal limits

The integral from a to a is F(a) − F(a) = 0. There is no width, so there are no strips to add, whatever f is. The integral of any function from 3 to 3 is 0.

A constant factor comes out

Multiply every height by a constant k and every strip $f\left(x_i\right)\Delta x$f(xᵢ) Δx becomes $kf\left(x_i\right)\Delta x$k f(xᵢ) Δx, so the whole sum is k times as big: $\int kf\left(x\right)dx=k\int f\left(x\right)dx$∫ k f(x) dx = k ∫ f(x) dx.

Check it: from 0 to 2, $\frac{x^2}{4}$x²/4 has integral $\frac{2}{3}$2/3. Twice the function, $\frac{x^2}{2}$x²/2, has antiderivative $\frac{x^3}{6}$x³/6 and integral $\frac{8}{6}=\frac{4}{3}$8/6 = 4/3, which is $2\times \frac{2}{3}$2 × 2/3. In the same way the integral of $3x^2$3x² from 0 to 2 is $2^3=8$2³ = 8, and 3 times the integral of $x^2$x², $3\times \frac{8}{3}$3 × 8/3, is 8 as well.

The gold curve $y=\frac{x^2}{4}$y = x²/4 and the dashed curve $y=\frac{x^2}{2}$y = x²/2, twice as tall at every x. The shaded region under the dashed curve from 0 to 2 has area $\frac{4}{3}$4/3, twice the $\frac{2}{3}$2/3 under the gold one.

A sum splits

At each x the height of f + g is the height of f plus the height of g, so each strip $\left(f\left(x_i\right)+g\left(x_i\right)\right)\Delta x$(f(xᵢ) + g(xᵢ)) Δx is $f\left(x_i\right)\Delta x+g\left(x_i\right)\Delta x$f(xᵢ) Δx + g(xᵢ) Δx. Adding up the strips, $\int \left(f+g\right)dx=\int fdx+\int gdx$∫ (f + g) dx = ∫ f dx + ∫ g dx.

Check it from 0 to 2 with $f\left(x\right)=x^2$f(x) = x² and g(x) = 2x. An antiderivative of $x^2+2x$x² + 2x is $\frac{x^3}{3}+x^2$x³/3 + x², which gives $\frac{8}{3}+4=\frac{20}{3}$8/3 + 4 = 20/3. Separately, $x^2$x² gives $\frac{8}{3}$8/3 and 2x gives $2^2=4$2² = 4, and $\frac{8}{3}+4=\frac{20}{3}$8/3 + 4 = 20/3.

A product does not split. From 0 to 1, the integral of $x\times x=x^2$x × x = x² is $\frac{1}{3}$1/3, but the integral of x is ½, and ½ × ½ = ¼.

Joining two intervals

The integral from a to b plus the integral from b to c is (F(b) − F(a)) + (F(c) − F(b)) = F(c) − F(a), the integral from a to c. The F(b) cancels.

On $y=\frac{x^2}{4}$y = x²/4: from 0 to 2 the integral is $\frac{2}{3}$2/3, from 2 to 4 it is $\frac{64}{12}-\frac{8}{12}=\frac{14}{3}$64/12 − 8/12 = 14/3, and $\frac{2}{3}+\frac{14}{3}=\frac{16}{3}$2/3 + 14/3 = 16/3, which is $\frac{64}{12}$64/12, the integral from 0 to 4.

The point in the middle need not lie between the other two, once a reversed integral counts negative. From 0 to 6 the integral is $\frac{216}{12}=18$216/12 = 18, and from 6 to 4 it is $-\left(18-\frac{16}{3}\right)=-\frac{38}{3}$−(18 − 16/3) = −38/3. Together they give $18-\frac{38}{3}=\frac{16}{3}$18 − 38/3 = 16/3, the integral from 0 to 4 again.

The region under $y=\frac{x^2}{4}$y = x²/4 from a = 0 to c = 4, with b = 2 marked on the axis. The part from a to b has area $\frac{2}{3}$2/3 and the part from b to c has area $\frac{14}{3}$14/3; together they make $\frac{16}{3}$16/3.

The usual mistakes

Keeping the sign when the limits are swapped. If the integral from 2 to 5 is 7, the integral from 5 to 2 is −7, not 7.

Giving f(3) for the integral from 3 to 3. A height times no width is 0.

Adding the constant instead of multiplying by it. The integral of 5f(x) is 5 times the integral of f(x), not the integral of f(x) plus 5.

Splitting a product as if it were a sum. From 0 to 1, x × x integrates to $\frac{1}{3}$1/3, not to ½ × ½.

Multiplying or subtracting the pieces of a joined interval. If 0 to 2 gives 3 and 2 to 5 gives 4, then 0 to 5 gives 3 + 4 = 7.

A boiler’s gas

In the application below, the gas burnt over a morning is split at a change of tariff. The two pieces add back to the whole, and reading the last stretch backwards gives the same amount with its sign changed.