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🇨🇩 Algebra and analysis

Antiderivatives and integrals

⏱ 30 minOfficial programme: CD-H4-ANA-03

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🎯 In this lesson, you will learn to

  • Find antiderivatives by several methods; definite integrals.
  • Use integrals for areas and volumes.

The lesson

Introduction

A farmer in the Ruzizi plain wants to know the exact area of a field bordered on one side by a curving river. A potter wants to compute the volume of a rounded jar. These calculations use the integral, which adds up infinitely many very thin slices. To compute it, we look for an antiderivative, that is, a function whose derivative is the given function.

Explanation

Antiderivatives

FF is an antiderivative of ff on an interval II if F′=fF'=f on II. The other antiderivatives are the functions F+CF+C, CC a real constant.

Function ffAntiderivative FF
xnx^n (n≠−1n\ne-1)xn+1n+1\dfrac{x^{n+1}}{n+1}
1x\dfrac1x (x>0x>0)ln⁡x\ln x
exe^xexe^x
cos⁡x\cos x; sin⁡x\sin xsin⁡x\sin x; −cos⁡x-\cos x
u′unu'u^n (n≠−1n\ne-1)un+1n+1\dfrac{u^{n+1}}{n+1}
u′u\dfrac{u'}{u} (u>0u>0)ln⁡u\ln u
u′euu'e^ueue^u

An antiderivative of a sum is the sum of the antiderivatives; a constant factor is kept: ∫kf=k∫f\int kf=k\int f.

Explanation

Definite integral and methods

If FF is an antiderivative of ff, continuous on [a,b][a, b]:
∫abf(x) dx=[F(x)]ab=F(b)−F(a).\int_a^b f(x)\,\mathrm{d}x=\big[F(x)\big]_a^b=F(b)-F(a).
Properties: linearity; Chasles’ relation ∫abf=∫acf+∫cbf\int_a^b f=\int_a^c f+\int_c^b f; if f≥0f\ge0 on [a,b][a, b] (with a<ba<b), then ∫abf≥0\int_a^b f\ge0.

Substitution: recognise a form u′ g(u)u'\,g(u), for example ∫2x ex2dx=ex2+C\int 2x\,e^{x^2}\mathrm{d}x=e^{x^2}+C.

Integration by parts:
∫abu v′ dx=[u v]ab−∫abu′ v dx.\int_a^b u\,v'\,\mathrm{d}x=\big[u\,v\big]_a^b-\int_a^b u'\,v\,\mathrm{d}x.
Choose as uu the part that becomes simpler when differentiated (a polynomial, ln⁡x\ln x…).

Explanation

Areas and volumes

  • If f≥0f\ge0 on [a,b][a, b], the area of the region bounded by the curve, the xx-axis and the lines x=ax=a, x=bx=b is ∫abf(x) dx\displaystyle\int_a^b f(x)\,\mathrm{d}x (in area units).
  • If f≥gf\ge g on [a,b][a, b], the area between the two curves is ∫ab(f(x)−g(x)) dx\displaystyle\int_a^b\big(f(x)-g(x)\big)\,\mathrm{d}x.
  • If ff changes sign, split the interval and take the absolute value of each piece.
  • Volume of revolution: rotating the graph of ff about the xx-axis on [a,b][a, b] gives a solid of volume
    V=π∫abf(x)2 dx.V=\pi\int_a^b f(x)^2\,\mathrm{d}x.

Worked example

Problem: Compute I=∫02(3x2−2x+1) dxI=\displaystyle\int_0^2(3x^2-2x+1)\,\mathrm{d}x and J=∫01xex dxJ=\displaystyle\int_0^1xe^x\,\mathrm{d}x.

Step 1: An antiderivative of 3x2−2x+13x^2-2x+1 is x3−x2+xx^3-x^2+x, so I=(8−4+2)−0=6I=(8-4+2)-0=6.

Step 2: For JJ, by parts with u=xu=x, v′=exv'=e^x: u′=1u'=1, v=exv=e^x.

Step 3: J=[xex]01−∫01ex dx=e−[ex]01=e−(e−1)=1J=\big[xe^x\big]_0^1-\displaystyle\int_0^1e^x\,\mathrm{d}x=e-\big[e^x\big]_0^1=e-(e-1)=1.

Answer: I=6I=6 and J=1J=1.

Worked example

Problem: Compute the area between the curves y=xy=x and y=x2y=x^2 on [0,1][0, 1].

Step 1: On [0,1][0, 1], x≥x2x\ge x^2 (for example 0.5≥0.250.5\ge0.25).

Step 2: A=∫01(x−x2) dx=[x22−x33]01\mathcal{A}=\displaystyle\int_0^1(x-x^2)\,\mathrm{d}x=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1.

Step 3: A=12−13=16\mathcal{A}=\dfrac12-\dfrac13=\dfrac16.

Answer: 16\dfrac16 area unit.

Worked example

Problem: The graph of f(x)=xf(x)=\sqrt x on [0,4][0, 4] is rotated about the xx-axis (unit: the decimetre). Compute the volume of the bowl-shaped solid obtained.

Step 1: f(x)2=xf(x)^2=x.

Step 2: V=π∫04x dx=π[x22]04=π×8V=\pi\displaystyle\int_0^4x\,\mathrm{d}x=\pi\left[\frac{x^2}{2}\right]_0^4=\pi\times8.

Answer: V=8π≈25.13V=8\pi\approx25.13 dm³, about 2525 litres.

Watch out

Watch out: An integral is not always an area: if ff is negative on part of the interval, that part counts negatively. For example ∫02πsin⁡x dx=0\displaystyle\int_0^{2\pi}\sin x\,\mathrm{d}x=0, whereas the area between the curve and the axis is 44. And remember to compute F(b)−F(a)F(b)-F(a) in that order, also evaluating F(a)F(a), even when a=0a=0: [ex]01=e−1\big[e^x\big]_0^1=e-1, not ee.

Key points

  • F′=fF'=f; antiderivatives of xnx^n, 1x\dfrac1x, exe^x, cos⁡x\cos x, sin⁡x\sin x, u′unu'u^n, u′u\dfrac{u'}u, u′euu'e^u.
  • ∫abf(x) dx=F(b)−F(a)\displaystyle\int_a^bf(x)\,\mathrm{d}x=F(b)-F(a).
  • By parts: ∫abuv′=[uv]ab−∫abu′v\displaystyle\int_a^buv'=\big[uv\big]_a^b-\int_a^bu'v.
  • Area between two curves: ∫ab(f−g)\displaystyle\int_a^b(f-g) with f≥gf\ge g.
  • Volume of revolution: V=π∫abf(x)2 dxV=\pi\displaystyle\int_a^bf(x)^2\,\mathrm{d}x.

✍️ Exercises

Answer one exercise at a time. Your answer is checked right away, with an explanation.

Exercise 1 of 9

★ Easy

Which function is an antiderivative of f(x)=6x2−4x+1f(x)=6x^2-4x+1 on R\mathbb{R}?

Your answer