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Calculus Clinic · Integration

Forgetting the constant of integration

Every indefinite integral needs a constant: f(x)dx=F(x)+c\int f(x)\,dx=F(x)+c. The error is dropping the +c+c.

It is harmless only until a boundary condition is given, then the missing cc is exactly the number you were meant to find.

The error

Integration reverses differentiation, and differentiation destroys constants: the derivative of x2+7x^{2}+7 and of x23x^{2}-3 is the same 2x2x. So when you integrate 2x2x you cannot know which constant was there, you must write +c+c to stand for every possibility.

The tempting mistake in Add Math is to leave it off, because the power-rule part feels like the whole answer and the +c+c looks like decoration. On a plain "integrate this" question it may only cost a mark.

But the moment the question gives a point on the curve, or a value the function must take, the +c+c becomes the unknown you are meant to solve for. Drop it and there is nothing left to substitute into, so the rest of the question collapses.

Wrong line, then the fix

Integrate 6x24x+16x^{2}-4x+1. Applying the power rule term by term but stopping there gives:

Wrong, the constant is missingMust memorise
(6x24x+1)dx=2x32x2+x\int (6x^{2}-4x+1)\,dx=2x^{3}-2x^{2}+x

An indefinite integral is a family of curves, not one curve, so the constant must be there:

CorrectMust memorise
(6x24x+1)dx=2x32x2+x+c\int (6x^{2}-4x+1)\,dx=2x^{3}-2x^{2}+x+c

Check by differentiating the right-hand side: ddx(2x3)=6x2\frac{d}{dx}(2x^{3})=6x^{2}, ddx(2x2)=4x\frac{d}{dx}(-2x^{2})=-4x, ddx(x)=1\frac{d}{dx}(x)=1, and ddx(c)=0\frac{d}{dx}(c)=0. The cc vanishes on the way back, which is precisely why it has to be written on the way forward.

The reliable fix

  1. As you integrate, write +c+c on the same line, before you do anything else, treat it as part of the integral sign, not an afterthought.
  2. Ask whether the question is indefinite (an integral with no limits) or definite (with limits). Only indefinite integrals carry +c+c; in a definite integral it cancels out.
  3. If a point or condition is given, substitute it into your F(x)+cF(x)+c to form an equation in cc.
  4. Solve that equation for cc, then write the final function with the specific value in place.
  5. Verify by differentiating your answer back and checking it also passes through the given point.

One line to remember it

No limits, no answer without +c

If the integral has no upper and lower limits, your line is unfinished until the +c+c is on it. The constant is not decoration, it is usually the thing the next line asks you to find.

Practice where the error hides

Q1[4 marks]

A curve has gradient function dydx=3x24x\frac{dy}{dx}=3x^{2}-4x and passes through the point (2,5)(2,\,5). Find the equation of the curve in terms of xx.

Show worked solution

To get yy from its gradient function, integrate, and keep the constant, because we will need it:

y=(3x24x)dx=x32x2+cy=\int (3x^{2}-4x)\,dx=x^{3}-2x^{2}+c

The curve passes through (2,5)(2,5), so substitute x=2x=2 and y=5y=5:

5=(2)32(2)2+c=88+c=c5=(2)^{3}-2(2)^{2}+c=8-8+c=c

Therefore c=5c=5, and the equation of the curve is:

y=x32x2+5y=x^{3}-2x^{2}+5

Check: at x=2x=2, y=88+5=5y=8-8+5=5, matching the given point, and differentiating gives 3x24x3x^{2}-4x as required. Had the +c+c been dropped, there would have been no unknown to fit the point (2,5)(2,5), and the answer would silently be wrong for every xx.

How one-to-one teaching helps

Our teachers build the +c+c into your hand: it goes down at the same moment the integral is written, not after. In a one-to-one lesson we practise the tell-tale phrase "passes through" or "when x=, y=x=\ldots,\ y=\ldots", which signals that the constant is the whole point of the question.

Because the paper awards method marks, keeping the constant and forming the equation in cc earns credit step by step even if the arithmetic wobbles. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To lock in the habit, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

When can I safely leave out the +c+c?

Only for a definite integral, one written with a lower and upper limit. There the constant appears at both limits and cancels when you subtract, so it never affects the value.

For any indefinite integral (no limits), the +c+c must be there.

Why is the constant sometimes the whole answer?

When a question gives a point on the curve or a starting value, the shape of F(x)F(x) is fixed but its vertical position is not, and cc sets that position. Substituting the given point produces an equation whose solution is cc, so the constant is literally what you are asked to find.

Do I lose marks for forgetting the +c+c?

Yes, under analytic marking you typically lose the mark reserved for it, and if a boundary condition follows, you lose the marks that depended on solving for cc as well. Writing +c+c as you integrate protects the whole chain of method marks.

How do I check I kept the right constant?

Differentiate your final answer: it should return the original gradient function, and the constant should differentiate to zero. Then substitute the given point to confirm the equation holds, for y=x32x2+5y=x^{3}-2x^{2}+5 at (2,5)(2,5), both checks pass.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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