The Antiderivative

The question integration asks

Differentiating starts from a function and asks for its gradient. Integrating starts from the gradient and asks for the function. If F'(x) = f(x), then F is called an antiderivative of f.

Every derivative already learned can be read backwards this way. $3x^2$3x² is the derivative of $x^3$x³, so $x^3$x³ is an antiderivative of $3x^2$3x². sinh x differentiates to cosh x, so sinh x is an antiderivative of cosh x, and cosh x is an antiderivative of sinh x. $e^x$eˣ is an antiderivative of itself, and sin x is an antiderivative of cos x.

Each of these claims is checked the same way: differentiate the proposed answer and see whether the original function comes back.

Last move first

The power rule makes two moves on $x^3$x³. First the power 3 comes down as a multiplier, then the power drops by one, giving $3x^2$3x². To undo a pair of moves, undo the second one first. So start from $3x^2$3x² and put one back on the power, giving $3x^3$3x³. Then undo the multiply by dividing by the new power, 3: $\frac{3x^3}{3}=x^3$3x³/3 = x³.

The order matters. Dividing by the old power first gives $\frac{3x^2}{2}$3x²/2, and then raising the power gives $\frac{3x^3}{2}$3x³/2. That differentiates to $\frac{9x^2}{2}$9x²/2, not $3x^2$3x². The divisor has to be the power the term ends up with, because that is the number differentiating brings down.

The gold curve is $F\left(x\right)=x^3$F(x) = x³ and the plain curve is $f\left(x\right)=3x^2$f(x) = 3x². At x = 1 the tangent to F, y = 3x − 2, has gradient 3, and f(1) = 3. At x = −1 the tangent to F, y = 3x + 2, also has gradient 3, and f(−1) = 3. The height of f at each x is the gradient of F there.

One answer gives all of them

If F'(x) = f(x), then F(x) + 5, F(x) − 2 and F(x) + C for any constant C are antiderivatives as well, because a constant differentiates to 0. On a graph, adding C slides the curve straight up or down, and sliding a curve does not change how steep it is at any x.

Nothing else works. Suppose F and G are both antiderivatives of f. Then G'(x) − F'(x) = f(x) − f(x) = 0, so the function G − F has gradient 0 everywhere. Along an interval, a function with gradient 0 never rises or falls, so it is a constant. Any two antiderivatives of the same function differ by a constant, and the complete answer is F(x) + C.

Two answers that look different

Take f(x) = 2x + 2. One person integrates term by term and gets $x^2+2x$x² + 2x. Another notices that ${\left(x+1\right)}^2$(x + 1)² differentiates, by the chain rule, to 2(x + 1) = 2x + 2, and gets ${\left(x+1\right)}^2$(x + 1)². Both differentiate to 2x + 2, so both are right.

They are not the same function: ${\left(x+1\right)}^2=x^2+2x+1$(x + 1)² = x² + 2x + 1, which is 1 more than $x^2+2x$x² + 2x at every x. That 1 is the constant between two members of one family. Written with + C, the two answers describe the same set of curves.

The gold curve is $y={\left(x+1\right)}^2$y = (x + 1)² and the plain curve is $y=x^2+2x$y = x² + 2x, exactly 1 below it everywhere. At x = 0.5 their tangents, y = 3x + 0.75 through (0.5, 2.25) and y = 3x − 0.25 through (0.5, 1.25), are parallel, with gradient 3.

The family $y=x^2+C$y = x² + C, with C = −1 picked out and its tangent at x = 1, gradient 2. The dashed tangent is the one on $y=x^2$y = x² at the same x, parallel to it. Drag C and the curve slides; drag x and the gradient follows 2x, whatever C is.

Check by differentiating

Suppose the claim is that $6x^2-4x+5$6x² − 4x + 5 integrates to $2x^3-2x^2+5x+C$2x³ − 2x² + 5x + C. Differentiate: $6x^2-4x+5$6x² − 4x + 5, the original. The claim stands. The same check confirms $\sinh x+x^2+C$sinh x + x² + C as the integral of cosh x + 2x.

The check also catches mistakes. A tempting way to integrate x(x + 1) is to integrate each factor, $\frac{x^2}{2}$x²/2 and $\frac{x^2}{2}+x$x²/2 + x, and multiply them, giving $\frac{x^4}{4}+\frac{x^3}{2}$x⁴/4 + x³/2. That differentiates to $x^3+\frac{3x^2}{2}$x³ + 3x²/2, which is 2.5 at x = 1, while x(x + 1) is 2 there. Multiply out first instead: $x\left(x+1\right)=x^2+x$x(x + 1) = x² + x, which integrates to $\frac{x^3}{3}+\frac{x^2}{2}+C$x³/3 + x²/2 + C, and that differentiates back to $x^2+x$x² + x.

The usual mistakes

Differentiating instead. The integral of $x^4$x⁴ is not $4x^3$4x³; integrating raises the power.

Raising the power without dividing. $x^4$x⁴ is not the integral of $x^3$x³, because it differentiates to $4x^3$4x³, four times too much. The integral of $x^3$x³ is $\frac{x^4}{4}$x⁴/4.

Dividing by the old power. $\frac{x^4}{3}$x⁴/3 differentiates to $\frac{4x^3}{3}$4x³/3, not $x^3$x³.

Leaving out the + C. Without it the answer is one curve of the family, and a later question that names a point on the curve has no constant left to find.