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P1.5 - Integration

Syllabus
2019
Topic
P1.5
Level
AS

Learning objectives

P1.5.1Indefinite integration as the reverseIndefinite integration as the reverse Students should know that a constant of integration is of differentiation. required.P1.5.2Integration of powers of x and related sumsIntegration of xn and related sums, (Excluding n = −1 and related sums, differences and differences and constant multiples. multiples).; For example, the ability to integrate expressions such as 1 x2 −3x −1 2 and (x + 2)2 is expected. x Given f ′(x) and a point on the curve, students should be able to find an equation of the curve in the form y = f(x).

Indefinite integration recovers a family of functions

Indefinite integration reverses differentiation. It finds every function whose derivative is the given integrand, so its result is a family rather than one unique curve.

F′(x)=f(x)⟺∫f(x) dx=F(x)+CF'(x)=f(x)\quad\Longleftrightarrow\quad\int f(x)\,dx=F(x)+C

The constant of integration CC is required because every constant differentiates to zero. Curves such as F(x)+2F(x)+2 and F(x)−7F(x)-7 have the same derivative, so derivative information alone cannot distinguish their vertical positions.

Since ddx(2x3−2x2+5x)=6x2−4x+5\dfrac{d}{dx}(2x^3-2x^2+5x)=6x^2-4x+5, ∫(6x2−4x+5) dx=2x3−2x2+5x+C.\int(6x^2-4x+5)\,dx=2x^3-2x^2+5x+C. Differentiating the result removes CC and reproduces the integrand, which is the most direct check.

Do not omit +C+C from an indefinite integral. A given point on a particular curve can later determine its value; without such information, leaving one arbitrary constant is essential.

Integrate powers, then use a point to locate the curve

∫xn dx=xn+1n+1+C,n≠−1\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1

For each term axnax^n, increase the power by one and divide the coefficient by that new power. First expand brackets and rewrite roots or variables in denominators as powers, then integrate sums, differences and constant multiples term by term.

Structure Rewrite before integrating
x\sqrt{x} x1/2x^{1/2}
1/xp1/x^p x−px^{-p}
(x+2)2/x(x+2)^2/\sqrt{x} x3/2+4x1/2+4x−1/2x^{3/2}+4x^{1/2}+4x^{-1/2}

Therefore ∫(x+2)2x dx=25x5/2+83x3/2+8x1/2+C.\int\frac{(x+2)^2}{\sqrt{x}}\,dx=\frac25x^{5/2}+\frac83x^{3/2}+8x^{1/2}+C. Differentiating each term returns the rewritten integrand.

If f′(x)=3x2−4f'(x)=3x^2-4 and the curve passes through (2,5)(2,5), then f(x)=x3−4x+Cf(x)=x^3-4x+C. Substitution gives 5=8−8+C5=8-8+C, so C=5C=5 and f(x)=x3−4x+5f(x)=x^3-4x+5.

The stated power rule excludes n=−1n=-1, because dividing by n+1n+1 would divide by zero; that case is outside this P1 scope. Preserve restrictions such as x>0x>0 after rewriting expressions containing x\sqrt{x} or denominators.