Losing the +c in Indefinite Integrals

Q: Why does an indefinite integral need a constant at all?

Because differentiation destroys constants, x4−3x2x^{4}-3x^{2}x4−3x2 and x4−3x2+2x^{4}-3x^{2}+2x4−3x2+2 have the same derivative. Reversing that step cannot know which constant was there, so +c+c+c stands for the whole family of curves that share the given gradient function.

Q: How does a given point fix the constant?

The shape of F(x)F(x)F(x) is set by the integration, but its vertical position is not, that is what ccc controls. Substituting the given point produces an equation whose solution is ccc, selecting the one curve in the family that passes through that point.

The error

Differentiation flattens constants: $x^{4}-3x^{2}$ , $x^{4}-3x^{2}+2$ and $x^{4}-3x^{2}-5$ all have the same derivative, because the constant differentiates to zero. Integration runs that in reverse, so when you integrate you cannot recover which constant was there, every possibility is still open.

That is what the $+c$ records: an indefinite integral names a whole family of curves stacked vertically, not a single one. The tempting mistake in Add Math is to write only $F(x)$ and stop, because the power-rule part looks like the finished answer and the $+c$ feels like decoration.

On a bare "integrate this" it may cost a single mark. But the moment a point or a starting value is given, the $+c$ is the one number that selects the correct curve from the family, leave it out and you have silently chosen the member through the origin, which is almost never the one the question wants.

Wrong line, then the fix

Integrate $4x^{3}-6x$ . Applying the power rule term by term but stopping there gives:

Wrong, this is only one member of the familyMust memorise

\int (4x^{3}-6x)\,dx=x^{4}-3x^{2}

An indefinite integral is the whole family, so the constant must be written:

CorrectMust memorise

\int (4x^{3}-6x)\,dx=x^{4}-3x^{2}+c

Every curve below shares the exact derivative $4x^{3}-6x$ ; only their heights differ. The $+c$ is what keeps all of them in play until the question tells you which one to keep.

Member of the family	Its derivative
y = x^4 - 3x^2	4x^3 - 6x
y = x^4 - 3x^2 + 2	4x^3 - 6x
y = x^4 - 3x^2 - 5	4x^3 - 6x

The reliable fix

Write $+c$ on the same line as you integrate, before anything else, treat it as part of finishing the integral, not an afterthought.
Ask whether the integral is indefinite (no limits) or definite (with limits). Only indefinite integrals carry $+c$ ; in a definite integral it cancels on subtraction.
If a point or starting value is given, substitute it into $F(x)+c$ to form an equation in $c$ .
Solve for $c$ , then rewrite the final answer with that specific value in place.
Check by differentiating your answer back to the original function, and confirm it passes through the given point.

One line to remember it

One integral, a whole family

An indefinite integral answers with a family of curves, not one line. Until the $+c$ is written, the family is incomplete, and the constant is usually the number the next line asks you to find.

Practice where the error hides

Q1[4 marks]

A curve has gradient function $\frac{dy}{dx}=8x-3$ and passes through the point $(2,\,7)$ . Find the equation of the curve.

Show worked solution

Integrate the gradient function to get $y$ , keeping the constant because the given point will pin it down:

y=\int (8x-3)\,dx=4x^{2}-3x+c

The curve passes through $(2,7)$ , so substitute $x=2$ and $y=7$ :

7=4(2)^{2}-3(2)+c=16-6+c=10+c

So $c=7-10=-3$ , and the equation of the curve is:

y=4x^{2}-3x-3

Check: at $x=2$ , $y=16-6-3=7$ , matching the given point, and differentiating gives $8x-3$ as required. Dropping the $+c$ would have left $y=4x^{2}-3x$ , which gives $y=10$ at $x=2$ , the wrong member of the family, missing the point by $3$ .

How one-to-one teaching helps

Our teachers build the $+c$ into your hand so it lands at the same instant the integral is written, never afterwards. In a one-to-one lesson we train you to hear the signal words "passes through" or "when $x=\ldots,\ y=\ldots$ ", which mean the constant is the whole point of the question, then form and solve the equation in $c$ step by step.

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

To make the habit automatic, 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

Why does an indefinite integral need a constant at all?

Because differentiation destroys constants, $x^{4}-3x^{2}$ and $x^{4}-3x^{2}+2$ have the same derivative. Reversing that step cannot know which constant was there, so $+c$ stands for the whole family of curves that share the given gradient function.

When can I leave the

+c

out?

Only in a definite integral, 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, with no limits, the $+c$ must be there.

How does a given point fix the constant?

The shape of $F(x)$ is set by the integration, but its vertical position is not, that is what $c$ controls. Substituting the given point produces an equation whose solution is $c$ , selecting the one curve in the family that passes through that point.

Do I lose marks for forgetting the

+c

Yes. Under analytic marking you typically lose the mark reserved for the constant, and if a point follows, you also lose the marks that depended on solving for $c$ .

Writing $+c$ as you integrate protects the whole chain of method marks.

Source:SRC-DSKP-EN

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

Losing the constant in indefinite integration

The error

Wrong line, then the fix

The reliable fix

One line to remember it

Practice where the error hides

How one-to-one teaching helps

Frequently asked questions

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