Reversing the Power Rule
The power rule says that \(\dfrac{d}{dx}x^{n}=nx^{n-1}\). Differentiation lowers the exponent by one and multiplies by the old exponent. To undo it, we raise the exponent by one and divide by the new exponent. A polynomial is a sum of constant multiples of powers of \(x\), so once we know how to integrate each power, linearity lets us integrate the whole polynomial one term at a time.
The restriction \(n\ne-1\) is necessary because the formula would divide by zero. The case \(n=-1\) is handled separately with the natural logarithm and does not arise in polynomials. Verification is immediate: differentiating \(\dfrac{x^{n+1}}{n+1}\) gives \(\dfrac{(n+1)x^{n}}{n+1}=x^{n}\).
Two facts make polynomial integration routine. First, constant multiples come outside: \(\int cf(x)\,dx=c\int f(x)\,dx\). Second, sums split: \(\int(f+g)\,dx=\int f\,dx+\int g\,dx\). Together, these are linearity. Each term produces its own arbitrary constant, but the sum of finitely many arbitrary constants is again an arbitrary constant, so we write only one \(C\) at the end.
Worked Example 1: A Quadratic
Find \(\displaystyle\int(6x^2-4x+7)\,dx\).
Integrate each term: \(\int6x^2dx=6\cdot\dfrac{x^3}{3}=2x^3\); \(\int-4x\,dx=-4\cdot\dfrac{x^2}{2}=-2x^2\); \(\int7\,dx=7x\). Therefore
Check: differentiating gives \(6x^2-4x+7\). At \(x=1.7\), the original integrand equals 17.54, and a numerical derivative of \(2x^3-2x^2+7x\) gives 17.54.
Worked Example 2: A Quartic with Fractional Coefficients
Find \(\displaystyle\int(3x^4-2x^3+x-9)\,dx\).
Termwise: \(3\cdot\dfrac{x^5}{5}\), \(-2\cdot\dfrac{x^4}{4}=-\dfrac{x^4}{2}\), \(\dfrac{x^2}{2}\), and \(-9x\). So
Notice that the term \(x\) is \(x^1\), so its antiderivative is \(\dfrac{x^2}{2}\). It is easy to forget that a bare \(x\) is still a power.
Worked Example 3: Expanding First
Find \(\displaystyle\int(x+2)(x-3)\,dx\).
The power rule applies to powers of \(x\), not to products of binomials, so expand first: \((x+2)(x-3)=x^2-x-6\). Then
Check: the derivative of the result is \(x^2-x-6\), which factors back to \((x+2)(x-3)\).
Worked Example 4: Choosing the Constant
A function \(F\) satisfies \(F'(x)=3x^2-2x+1\) and \(F(1)=4\). Find \(F\).
Antidifferentiating, \(F(x)=x^3-x^2+x+C\). The condition gives \(1-1+1+C=4\), so \(C=3\). Hence
Check: \(F(1)=1-1+1+3=4\) and \(F'(x)=3x^2-2x+1\). Both conditions are met. The initial condition selects one curve from the family of all antiderivatives, which differ only by vertical shifts.
Worked Example 5: Fractional and Negative Powers
Find \(\displaystyle\int\left(\sqrt x+\frac1{x^2}\right)dx\) for \(x>0\).
Rewrite as powers: \(\sqrt x=x^{1/2}\) and \(1/x^2=x^{-2}\). Then \(\int x^{1/2}dx=\tfrac23x^{3/2}\) and \(\int x^{-2}dx=\dfrac{x^{-1}}{-1}=-\dfrac1x\). So
The power rule works for any exponent except \(-1\), including fractions and negative numbers.
Summary of the Procedure
| Step | Action | Example 1 |
|---|---|---|
| 1 | Expand or rewrite so the integrand is a sum of powers | 6x² − 4x + 7 |
| 2 | Raise each exponent by one | x³, x², x¹ |
| 3 | Divide each term by its new exponent | 6x³/3, −4x²/2, 7x |
| 4 | Simplify the coefficients | 2x³ − 2x² + 7x |
| 5 | Add + C and check by differentiating | + C |
Common Mistakes
Forgetting the constant. Writing \(\int2x\,dx=x^2\) is incomplete. The antiderivatives of \(2x\) are \(x^2+C\) for every constant \(C\); for instance \(x^2+5\) and \(x^2-1\) both differentiate to \(2x\).
Using \(x^{n}/n\) instead of \(x^{n+1}/(n+1)\). For \(\int x^3dx\), the correct answer is \(\tfrac{x^4}{4}+C\), not \(\tfrac{x^3}{3}+C\). Differentiating the wrong answer gives \(x^2\), not \(x^3\).
Treating a constant as zero. The antiderivative of the constant \(7\) is \(7x\), not \(0\) or \(7\). The graph of the antiderivative of a nonzero constant is a line with that constant as its slope.
Integrating a product factor by factor. It is false that \(\int(x+2)(x-3)\,dx\) equals \(\int(x+2)\,dx\cdot\int(x-3)\,dx\). Expand first, as in Example 3.
Why Only One Constant
If \(F\) and \(G\) are both antiderivatives of \(f\) on an interval, then \((F-G)'=f-f=0\), and a function with derivative zero on an interval is constant. So \(F-G=C\) for some constant. This is why the family \(F(x)+C\) contains every antiderivative, and why a polynomial needs only one \(C\) even though each term could carry its own: \(C_1+C_2+C_3\) is just another constant. Graphically, the antiderivatives of \(6x^2-4x+7\) are copies of the same curve \(2x^3-2x^2+7x\) shifted up or down, and all of them have the same slope at each value of \(x\).
Worked Example 6: From Velocity to Position
A particle moves along a line with velocity \(v(t)=6t^2-4t\) (meters per second) and position \(s(0)=2\). Find \(s(t)\) and \(s(2)\).
Position is an antiderivative of velocity: \(s(t)=2t^3-2t^2+C\). The condition \(s(0)=2\) gives \(C=2\), so
At \(t=2\), \(s(2)=16-8+2=10\) meters. Check: \(s'(t)=6t^2-4t=v(t)\), and \(s(0)=2\). The same procedure, integrating a polynomial and fixing the constant from a starting value, applies to any quantity whose rate of change is known.
Checking by Differentiation
The most reliable habit in integration is to differentiate your answer. The power rule makes the check quick for polynomials: multiply each coefficient by its exponent and lower the exponent by one. For \(2x^3-2x^2+7x\), this returns \(6x^2-4x+7\). If the result differs from the integrand, the error is almost always one of three things: a wrong denominator after raising the exponent, a dropped coefficient, or an arithmetic slip when simplifying a fraction like \(3\cdot\tfrac15\). Because the check is mechanical, there is no reason to leave an indefinite integral unverified.
A related habit is to read the answer for plausibility. The degree of an antiderivative of a polynomial is one higher than the degree of the polynomial. A quadratic integrand gives a cubic, a quartic gives a quintic, and a constant gives a line. If the degree of your answer is not one more than the degree of the integrand, something went wrong in the exponent step.
Finally, notice how this tutorial sets up what follows. The next tutorial states linearity of indefinite integration formally and uses it for sums and constant multiples of more general functions, not only powers. Everything there rests on the habit practiced here: break the integrand into simple pieces, integrate each piece with a known rule, combine the results, and verify by differentiation. The polynomial case is the cleanest place to build that habit because every rule needed is a single power-rule step, and every answer can be checked in seconds.
Where This Fits in the Course
This tutorial belongs to the module “Antiderivatives and Basic Integrals”. The tutorial just before it is “Verifying The Power Rule By Differentiation,” and the one that follows is “Linearity Of Indefinite Integration.”
Practice Problems with Answers
Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.
- Find \(\int(8x^3-6x+1)\,dx\).
Answer. \(2x^4-3x^2+x+C\). - Find \(\int\left(\tfrac12x^2+5\right)dx\).
Answer. \(\tfrac{x^3}{6}+5x+C\). - Find \(\int(2x-1)^2\,dx\) by expanding.
Answer. \(\int(4x^2-4x+1)dx=\tfrac43x^3-2x^2+x+C\). - Find \(F\) with \(F'(x)=4x-3\) and \(F(2)=5\).
Answer. \(F=2x^2-3x+C\), \(8-6+C=5\) gives \(C=3\); \(F=2x^2-3x+3\).
Check Your Understanding
Work these without looking back.
- State the antiderivative of \(x^n\) and the one value of \(n\) it excludes.
- Find \(\int(5x^4-3x^2+2)\,dx\) and check by differentiating.
- Why does \(\int(x+1)(x+2)\,dx\) need to be expanded first?
- Find the antiderivative of \(6x^2+2\) whose graph passes through \((0,5)\).
- Explain why two antiderivatives of the same function differ only by a constant.
