Exercises — Antiderivative — definition, family of solutions (+C)

Power Rule with $n=5$: raise the power by one to $6$, divide by that new power. $\int x^{5}dx=\frac{x^6}{6}+C.$ Check (differentiate forward): $\frac{d}{dx}\left(\frac{x^6}{6}\right)=\frac{6x^5}{6}=x^5.✓$

Both are "known-slope" facts read backwards. Since $\frac{d}{dx}\sin x=\cos x$: $\int \cos xdx=\sin x+C$. Since $\frac{d}{dx}{e}^x=e^x$ (the exponential is its own slope): $\int e^{x}dx=e^x+C$.

Which function has slope $1$ everywhere? A straight line of slope $1$: $x$. $\int 1dx=x+C,\frac{d}{dx}(x+C)=1.✓$ (This is Power Rule with $n=0$: $x^0=1\to \frac{x^1}{1}=x$.)

Integrate term by term (the integral of a sum is the sum of the integrals; constants factor out). $\int 3x^{2}dx=3\cdot \frac{x^3}{3}=x^3,\int (-4x)dx=-4\cdot \frac{x^2}{2}=-2x^2,\int 7dx=7x.$ $\int \left(3x^2-4x+7\right)dx=x^3-2x^2+7x+C.$ Check: $\frac{d}{dx}(x^3-2x^2+7x+C)=3x^2-4x+7.✓$ (One $+C$ for the whole expression, not one per term.)

Rewrite the root as a power: $\sqrt{x}=x^{1/2}$. Then Power Rule with $n=\frac{1}{2}$. $\int x^{1/2}dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}{x}^{3/2}+C.$ Check: $\frac{d}{dx}\left(\frac{2}{3}{x}^{3/2}\right)=\frac{2}{3}\cdot \frac{3}{2}{x}^{1/2}=x^{1/2}=\sqrt{x}.✓$

Rewrite as a negative power: $\frac{5}{x^2}=5x^{-2}$, so $n=-2$ (which is not $-1$, so Power Rule is legal). $\int 5x^{-2}dx=5\cdot \frac{x^{-1}}{-1}+C=-\frac{5}{x}+C.$ Check: $\frac{d}{dx}\left(-5x^{-1}\right)=-5\cdot (-1)x^{-2}=5x^{-2}=\frac{5}{x^2}.✓$

$\frac{1}{x}=x^{-1}$, so $n=-1$. The Power Rule divides by $n+1=0$ — division by zero, forbidden. This is exactly the one exception. The correct antiderivative is $\int \frac{1}{x}dx=\ln |x|+C,$ valid on $x\ne 0$. The absolute value covers both signs: $\frac{d}{dx}\ln |x|=\frac{1}{x}$ for every $x\ne 0$. See Logarithmic & Exponential Integrals.

No. The "two antiderivatives differ by a constant" theorem needs one connected interval, because it relies on the Mean Value Theorem (zero slope on a single interval ⇒ flat). The set $x\ne 0$ is two intervals $(-\infty ,0)$ and $(0,\infty )$ with a gap. Each piece can sit at its own height: $F(x)=\{\begin{array}{ll}\ln (-x)+C_1,& x<0,\\ \ln (x)+C_2,& x>0,\end{array}$ with $C_1,C_2$ independent. Differentiating each branch gives $\frac{1}{x}$, yet $C_1\ne C_2$ is allowed. So $\ln |x|+C$ misses solutions with a jump across the gap.

On the single interval $(-\infty ,\infty )$: zero slope everywhere ⇒ $g(x)=C$, one constant (MVT). On $x\ne 0$ (two intervals): $h$ is constant on each piece but the constants need not match: $h(x)=\{\begin{array}{ll}{C}_1,& x<0,\\ C_2,& x>0.\end{array}$

Step 1 — general family. $\int (6x-2)dx=3x^2-2x+C.$ Step 2 — use the condition to pick one curve. The point $(1,4)$ must lie on it: $F(1)=3(1{)}^2-2(1)+C=3-2+C=1+C=4\Rightarrow C=3.$ $\boxed{F(x)=3x^2-2x+3.}$ Check: $F^{\prime }(x)=6x-2$ and $F(1)=3-2+3=4.✓$

Position is an antiderivative of velocity (velocity is the slope of position). Integrate: $s(t)=\int (t^2-4t)dt=\frac{t^3}{3}-2t^2+C.$ Apply $s(0)=5$: $\frac{0}{3}-0+C=5\Rightarrow C=5$. $\boxed{s(t)=\frac{t^3}{3}-2t^2+5.}$ Sanity check at a point: $s(3)=\frac{27}{3}-2(9)+5=9-18+5=-4$ m, and $s^{\prime }(t)=t^2-4t=v(t).✓$

$\int (\cos x+e^x)dx=\sin x+e^x+C.$ Condition: $F(0)=\sin 0+e^0+C=0+1+C=0\Rightarrow C=-1$. $\boxed{F(x)=\sin x+e^x-1.}$ Check: $F^{\prime }(x)=\cos x+e^x$, $F(0)=0+1-1=0.✓$

An antiderivative of $2x$ is $x^2+C$. By the Fundamental Theorem of Calculus, evaluate $[x^2+C]$ from $1$ to $3$: $(3^2+C)-(1^2+C)=9+C-1-C=8.$ The $+C$ cancels ($C-C=0$), so the answer is the plain number $8$ — this is why definite integrals never carry a constant.

Let $H(x)=G(x)-F(x)$. Then $H^{\prime }(x)=G^{\prime }(x)-F^{\prime }(x)=0$ for all $x\in I$. A function with zero slope on a single interval cannot rise or fall, so it is constant — this is precisely what the Mean Value Theorem guarantees. Hence $H(x)=C$, i.e. $G(x)=F(x)+C.✓$

Note $|x{|}^3=|x{|}^3$; since $|x{|}^2=x^2$ and cubing keeps the sign of $x$, we have $|x{|}^3=x^2\cdot |x|$. Split by sign:

• For $x>0$: $|x{|}^3=x^3$, so $F(x)=\frac{x^3}{3}$, $F^{\prime }(x)=x^2.✓$
• For $x<0$: $|x|=-x$, so $|x{|}^3=(-x{)}^3=-x^3$, giving $F(x)=-\frac{x^3}{3}$ and $F^{\prime }(x)=-x^2\ne x^2.$ ✗

So $F$ is not an antiderivative of $x^2$ on $x<0$. The genuine general antiderivative on all of $\mathbb{R}$ is the simpler $\frac{x^3}{3}+C$ (check: derivative $x^2$ for every sign of $x$, including $x=0$).

Since $\frac{d}{dx}\tan x={\sec }^{2}x$, both $\tan x+C$ and $\tan x+3$ have derivative ${\sec }^{2}x$, so each is a valid antiderivative. But only $\tan x+C$ is the general answer (the whole family); $\tan x+3$ is one particular member ($C=3$). Writing a specific number instead of $+C$ silently discards infinitely many solutions.