Constant Of Integration

(a) $f(x)=2x^3-2x^2+5x-2$

(b) $f(x)=\ln x+x^2+3$

(c) $f(\theta )=2\sin \theta -\tan \theta +5$

(d) $f(x)=4\arcsin x-{\displaystyle \frac{2\pi }{3}}$

In every part the method is the same: integrate to get the general antiderivative $F(x)+C$, then substitute the given point to solve for $C$.

Imposing $f(1)=3$:

so $f(x)=2x^3-2x^2+5x-2$.

Imposing $f(1)=4$ and recalling $\ln 1=0$:

so $f(x)=\ln x+x^2+3$.

Since $\sin 0=0$ and $\tan 0=0$, the condition $f(0)=5$ gives $C=5$ directly, so

Here the condition is imposed at $x=\frac{1}{2}$, where $\arcsin \frac{1}{2}={\displaystyle \frac{\pi }{6}}$:

so $f(x)=4\arcsin x-{\displaystyle \frac{2\pi }{3}}$. Notice that the initial condition need not be given at $x=0$.

$f(x)=x^4-x^3+x-2$

Two integrations are needed, and each one introduces its own constant. Determine the first constant before integrating again.

The condition f'(0) = 1 gives $0-0+C=1$, so $C=1$ and

The condition $f(0)=-2$ gives $D=-2$, so

$f(x)=-\sin x-\cos x+3x+4$

Since $\cos 0=1$ and $\sin 0=0$, the condition f'(0) = 2 gives

so f'(x) = -\cos x+\sin x+3.

The condition $f(0)=3$ gives

so $f(x)=-\sin x-\cos x+3x+4$.

$f(x)=x^3+x^2-4x+3$

Here both conditions are values of $f$ itself, so neither constant can be found until after the second integration. Carry both constants along.

From $f(1)=1$,

Therefore $f(x)=x^3+x^2-4x+3$.

(a) $y=2x^2-3x+3$

(b) $y=-{\displaystyle \frac{1}{x}}+3$

(c) $y=2x^{3/2}-6$

The slope at every point is ${\displaystyle \frac{dy}{dx}}$, so integrating the given slope produces the whole family of curves. The point then selects one member of that family.

Substituting $(2,5)$:

so $y=2x^2-3x+3$.

Substituting $(1,2)$ gives $-1+C=2$, so $C=3$ and

Substituting $(4,10)$, and using $4^{3/2}=8$:

so $y=2x^{3/2}-6$.

(a) $y=x^3-x^2+2$

(b) Every member has slope $3x^2-2x$, and two members differ by the constant $C_1-C_2$.

(c) No. The point $(1,0)$ forces $C=0$, and that curve passes through $(2,4)$, not $(2,5)$.

so the curve is $y=x^3-x^2+2$.

So at any fixed $x$ the two curves have the same slope, meaning they are parallel in direction. Subtracting,

which does not depend on $x$. This is exactly the vertical shift described in the section: the curves are congruent, one lying a fixed distance above the other.

so the only candidate is $y=x^3-x^2$. At $x=2$ this gives $8-4=4$, not $5$. Hence no member passes through both points. In general one point determines $C$ completely, so a second point is one condition too many.

(a) The second integration needs a new constant: $f(x)=x^3+Cx+D$.

(b) The constant was found but never inserted: $f(x)=x^2+x+3$.

(c) The condition must be applied to $f$, not to f': $C=1$, so $f(x)=x^4+1$.

with $C$ and $D$ independent. Two conditions are then required to pin both of them down.

Now $f(0)=0+0+C=C$, and the condition $f(0)=3$ gives $C=3$. Hence

so $f(x)=x^4+1$.

(a) $f(x)=\ln x-1$

(b) $f(x)=\arctan x-{\displaystyle \frac{\pi }{4}}$

so $f(x)=\ln x-1$.

so $f(x)=\arctan x-{\displaystyle \frac{\pi }{4}}$.

In both parts the point chosen was one where the antiderivative takes a value we happen to know exactly. That is the only reason such points are convenient; any point in the domain would do.

$f(x)=x^4+x^2-x$

Three integrations, three constants, and three conditions. Determine each constant as soon as its condition becomes usable.

From f''(0) = 2 we get $A=2$, so f''(x) = 12x^{2}+2.

From f'(0) = -1 we get $B=-1$, so f'(x) = 4x^{3}+2x-1.

From $f(0)=0$ we get $D=0$. Therefore

In general, recovering $f$ from its $n$-th derivative requires $n$ conditions.

(a) F'(x) = G'(x) = \sin 2x.

(b) $F(x)-G(x)={\displaystyle \frac{1}{2}}$.

(c) ${\sin }^{2}x+1$ and $-\frac{1}{2}\cos 2x+\frac{3}{2}$, which are the same function.

The difference is constant, exactly as the theory predicts: two antiderivatives of the same function on an interval can differ only by a constant.

The lesson is that two correct answers to the same integration problem may look different while describing the same family of curves.

(a) $y=x^3+2$

(b) $y=x^3$

(c) The vertical distance is $2$ for every $x$, because the two curves differ only by the constant of integration.

Substituting $(1,3)$ gives $1+C=3$, so $C=2$ and $y=x^3+2$.

The $x^3$ terms cancel because the two curves share the same antiderivative and differ only in $C$. This is the geometric meaning of the constant of integration: changing $C$ slides the entire curve up or down without bending it, so the gap between two members is the same at every $x$.

F'(x) = \dfrac{1}{x} on both pieces, because a constant differentiates to zero regardless of which constant it is. The result that two antiderivatives differ by a single constant requires an interval; the set $x\ne 0$ is two separate intervals, so each piece carries its own independent constant.

Differentiate on each piece separately. For ,

and for , writing $|x|=-x$,

So F'(x) = \dfrac{1}{x} at every point where $F$ is defined, even though $F$ is not of the form $\ln |x|+C$ for any single number $C$.

The reason is that the theorem behind the constant of integration, that two functions with the same derivative on an interval differ by a constant, is a consequence of the Mean Value Theorem, and the Mean Value Theorem needs the two points it compares to be joined by a path inside the domain. The set $\{x\ne 0\}$ is not an interval; it is the union of the two separate intervals $(-\mathrm{\infty },0)$ and $(0,\mathrm{\infty })$. Nothing connects them, so the constant on one side is free of the constant on the other.

This is why the formulas in this chapter are always stated on an interval, and why writing ${\displaystyle \int \frac{dx}{x}=\ln |x|+C}$ is understood to mean "on any interval not containing $0$". In practice the initial condition tells you which side you are on, and that is the side you work with.