P1 5.1, P1 5.2 - Integrating powers

P1 5.1, P1 5.2 - Integrating powers

Reverse power differentiation to find indefinite integrals. Understand why every general answer needs an arbitrary constant, use the integration power rule with its exception, and rewrite products and root denominators before integrating.

Reversing differentiation and the arbitrary constant

If a curve's gradient function is 2x2x, must the curve be y=x2y=x^2? Consider y=x2+4y=x^2+4 and y=x2−7y=x^2-7: both also differentiate to 2x2x. Differentiation has lost the constant height shift.

Antiderivative and indefinite integral

An antiderivative of f(x)f(x) is a function whose derivative is f(x)f(x). Indefinite integration finds the family of antiderivatives, written

∫f(x) dx.\int f(x)\,\mathrm{d}x.

Here f(x)f(x) is the expression being integrated and dx\mathrm{d}x identifies the variable.

For f(x)=2xf(x)=2x, the whole family is

∫2x dx=x2+C.\int2x\,\mathrm{d}x=x^2+C.

The constant of integration CC is arbitrary: it may be any real number. Each choice gives one curve, and differentiating any of them removes CC.

[DIAGRAM: asset_name: One gradient function and many curves; asset_slug: constant-family; caption: Vertical shifts change the curve's height but leave its tangent gradients unchanged.; description: The three curves y equals x squared minus two, y equals x squared, and y equals x squared plus two are shown with parallel tangents at x equals one. Their marked points have heights negative one, one and three; each tangent gradient is two.]
Vertical shifts change the curve's height but leave its tangent gradients unchanged.

The curves in the picture have different heights at x=1x=1, but their tangents there all have gradient 22. This is the information a derivative preserves and a vertical shift changes.

Key idea

x2x^2 is one antiderivative of 2x2x; x2+Cx^2+C is the general antiderivative. Without extra information, there is no reason to choose C=0C=0.

The arbitrary constant belongs to the final function, not to the integrand. There is no remaining integral sign or dx\mathrm{d}x in the result. If several terms are integrated separately, their constants can be combined into one CC. Also, ∫0 dx=C\int0\,\mathrm{d}x=C: a zero gradient allows any constant height.

Choose the family that includes every vertical shift.

Power rule with n not equal to minus one

To reverse differentiation, increase a power by one and divide by that new power.

Since d(x3)/dx=3x2\mathrm{d}(x^3)/\mathrm{d}x=3x^2, an antiderivative of x2x^2 is x3/3x^3/3. Dividing by 33 cancels the multiplier that differentiation will introduce. More generally, differentiating xn+1/(n+1)x^{n+1}/(n+1) gives xnx^n, provided n+1≠0n+1\ne0.

Power rule for integration

For n≠−1n\ne-1, on an interval where the powers are real and differentiable,

∫kxn dx=kn+1xn+1+C.\int kx^n\,\mathrm{d}x=\frac{k}{n+1}x^{n+1}+C.

Integrate sums and differences term by term, with one constant at the end. In particular, ∫k dx=kx+C\int k\,\mathrm{d}x=kx+C.

Not in the formula booklet: learn it.

The rule includes negative and fractional powers. For example, ∫x−1/2 dx=2x1/2+C\int x^{-1/2}\,\mathrm{d}x=2x^{1/2}+C for x>0x>0. The excluded power −1-1 would require division by zero; this power rule cannot integrate 1/x1/x.

Worked example 1: integrate a sum

Question. Find ∫(6x2−4x+5) dx\int(6x^2-4x+5)\,\mathrm{d}x.

Plan. Increase each exponent by one, divide its coefficient by the new exponent, and include CC.

∫(6x2−4x+5) dx=63x3−42x2+5x+C=2x3−2x2+5x+C\begin{aligned} \int(6x^2-4x+5)\,\mathrm{d}x &=\frac63x^3-\frac42x^2+5x+C\\ &=2x^3-2x^2+5x+C \end{aligned}

Answer. 2x3−2x2+5x+C2x^3-2x^2+5x+C.

Check. Differentiating gives 6x2−4x+5+06x^2-4x+5+0, exactly the integrand.

Change only the first exponent from 22 to −2-2. The new exponent is −1-1, so its coefficient is divided by a negative number.

Worked example 2: a negative power

Question. Find ∫(6x−2−4x+5) dx\int(6x^{-2}-4x+5)\,\mathrm{d}x, for x>0x>0.

Plan. Apply the same power rule; the first exponent changes from −2-2 to −1-1.

∫(6x−2−4x+5) dx=6−1x−1−42x2+5x+C=−6x−1−2x2+5x+C\begin{aligned} \int(6x^{-2}-4x+5)\,\mathrm{d}x &=\frac6{-1}x^{-1}-\frac42x^2+5x+C\\ &=-6x^{-1}-2x^2+5x+C \end{aligned}

Answer. −6x−2x2+5x+C-\frac6x-2x^2+5x+C.

Check. The derivative of −6x−1-6x^{-1} is (−6)(−1)x−2=6x−2(-6)(-1)x^{-2}=6x^{-2}; the remaining terms give −4x+5-4x+5.

An integrand with exponent −2-2 is allowed; an integrand with exponent −1-1 is the exception. Do not confuse the exponent before integration with the exponent after it.

Watch out

∫6x2 dx=2x3+C\int6x^2\,\mathrm{d}x=2x^3+C, not 18x3+C18x^3+C: integration divides by the new exponent. A constant such as 55 integrates to 5x5x; it does not disappear.

Change the first coefficient and check its sign carefully.

Rewriting before integration

The power rule works on separate powers, so expand and divide before using it on a product or a quotient.

You already used this strategy for differentiation. For example, (x+2)2=x2+4x+4(x+2)^2=x^2+4x+4, while division by x=x1/2\sqrt{x}=x^{1/2} reduces each of those exponents by 1/21/2. Rewriting changes the form, not the function.

Problem-solving

Before integrating, aim for a sum of terms kxnkx^n. Expand a bracketed numerator, split a single-power denominator across every term, and simplify the indices. Then check that none of the resulting powers is −1-1.

Start with the squared bracket on its own.

Worked example 3: expand before integrating

Question. Find ∫(x+2)2 dx\int(x+2)^2\,\mathrm{d}x.

Plan. Expand the square, then integrate the three polynomial terms.

∫(x+2)2 dx=∫(x2+4x+4) dx=x33+2x2+4x+C\begin{aligned} \int(x+2)^2\,\mathrm{d}x &=\int(x^2+4x+4)\,\mathrm{d}x\\ &=\frac{x^3}{3}+2x^2+4x+C \end{aligned}

Answer. 13x3+2x2+4x+C\frac13x^3+2x^2+4x+C.

Check. Differentiating gives x2+4x+4=(x+2)2x^2+4x+4=(x+2)^2, matching the original unexpanded integrand.

Change only the denominator: divide the same square by x\sqrt{x}. This introduces fractional powers, but not a new integration rule.

Worked example 4: a root denominator

Question. Find ∫(x+2)2x dx\int\frac{(x+2)^2}{\sqrt{x}}\,\mathrm{d}x, for x>0x>0.

Plan. Expand, divide every term by x1/2x^{1/2}, and integrate the resulting powers.

(x+2)2x=x2+4x+4x1/2=x3/2+4x1/2+4x−1/2\begin{aligned} \frac{(x+2)^2}{\sqrt{x}} &=\frac{x^2+4x+4}{x^{1/2}}\\ &=x^{3/2}+4x^{1/2}+4x^{-1/2} \end{aligned}

Thus

∫(x+2)2x dx=x5/25/2+4x3/23/2+4x1/21/2+C=25x5/2+83x3/2+8x1/2+C\begin{aligned} \int\frac{(x+2)^2}{\sqrt{x}}\,\mathrm{d}x &=\frac{x^{5/2}}{5/2} +\frac{4x^{3/2}}{3/2} +\frac{4x^{1/2}}{1/2}+C\\ &=\frac25x^{5/2}+\frac83x^{3/2}+8x^{1/2}+C \end{aligned}

Why does the last coefficient become 88? Dividing 44 by 1/21/2 doubles it.

Answer. 25x5/2+83x3/2+8x+C\frac25x^{5/2}+\frac83x^{3/2}+8\sqrt{x}+C.

Check. Differentiating the answer gives x3/2+4x1/2+4x−1/2x^{3/2}+4x^{1/2}+4x^{-1/2}. Multiplying this by x\sqrt{x} gives x2+4x+4x^2+4x+4, confirming the original quotient.

The opening question has a family of answers: every y=x2+Cy=x^2+C has gradient 2x2x. The same arbitrary constant is needed after more complicated integrations too.

Watch out

∫A(x)B(x) dx\int\frac{A(x)}{B(x)}\,\mathrm{d}x is not found by integrating numerator and denominator separately and dividing. Here the useful move was algebraic division into powers before integration.

Use the same expansion-and-division method with a new bracket.

Quick Recap

  • Indefinite integration reverses differentiation and gives a family of functions.
  • The constant CC records the vertical shift that differentiation cannot detect.
  • Increase each exponent by one and divide by that new exponent; this rule excludes an integrand power of −1-1.
  • Rewrite products and root denominators as sums of powers first, then differentiate your answer to check it.