Adding up over an inter­val

The def­i­nite inte­gral ∫abf(x) dx\displaystyle \int_a^b f(x)\,dx adds up ff over a whole inter­val. On a graph, it is the signed area under the curve. The beau­ti­ful sur­prise, called the fun­da­men­tal the­o­rem of cal­cu­lus, is that you do not need to add up any­thing by hand: any anti­deriv­a­tive gives you the answer. We will eval­u­ate def­i­nite inte­grals, includ­ing by sub­sti­tu­tion, learn the prop­er­ties that turn long prob­lems into short ones, and fin­ish with mixed prac­tice.

Def­i­nite inte­grals

For ff con­tin­u­ous on [a,b][a, b], pic­ture the area under the curve filled with thin rec­tan­gles; the def­i­nite inte­gral is the limit of their total area as they get thin­ner. The fun­da­men­tal the­o­rem of cal­cu­lus then says: if F′=fF' = f, then

∫abf(x) dx=F(b)−F(a)=[F(x)]ab.\displaystyle \int_a^b f(x)\,dx = F(b) - F(a) = \big[F(x)\big]_a^b.

Exam­ple 1. ∫13(2x2−x) dx=[2x33−x22]13=(18−4.5)−(23−12)=403\displaystyle \int_1^3(2x^2 - x)\,dx = \left[\tfrac{2x^3}{3} - \tfrac{x^2}{2}\right]_1^3 = (18 - 4.5) - \left(\tfrac{2}{3} - \tfrac{1}{2}\right) = \tfrac{40}{3}.

The curve y = 2x squared - x with the region under it from x = 1 to x = 3 shaded and filled with ten thin rectangles; the exact area is 40/3.
The inte­gral from 1 to 3 is the limit of the rec­tan­gle sums: 40/3.

Exam­ple 2. ∫0π/4sec⁡2x dx=tan⁡π4−tan⁡0=1\displaystyle \int_0^{\pi/4}\sec^2 x\,dx = \tan\tfrac{\pi}{4} - \tan 0 = 1. ∫01dx1+x2=π4\displaystyle \int_0^1\frac{dx}{1 + x^2} = \tfrac{\pi}{4}.

With sub­sti­tu­tion, remem­ber to change the lim­its as well, and you never need to go back to the old vari­able. Exam­ple 3. ∫02xx2+5 dx\displaystyle \int_0^2 x\sqrt{x^2 + 5}\,dx. As we move across the inter­val, t=x2+5t = x^2 + 5 runs from 55 to 99, so we get 12∫59t1/2 dt=13(27−55)\displaystyle \tfrac{1}{2}\int_5^9 t^{1/2}\,dt = \tfrac{1}{3}\left(27 - 5\sqrt{5}\right).

Exam­ple 4. ∫0π/2sin⁡2xcos⁡x dx=[sin⁡3x3]0π/2=13\displaystyle \int_0^{\pi/2}\sin^2 x\cos x\,dx = \left[\tfrac{\sin^3 x}{3}\right]_0^{\pi/2} = \tfrac{1}{3}.

Prop­er­ties that save work

  1. ∫abf=−∫baf\displaystyle \int_a^b f = -\int_b^a f; ∫aaf=0\displaystyle \int_a^a f = 0.
  2. ∫abf=∫acf+∫cbf\displaystyle \int_a^b f = \int_a^c f + \int_c^b f.
  3. ∫abf(x) dx=∫abf(a+b−x) dx\displaystyle \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx.
  4. ∫0af(x) dx=∫0af(a−x) dx\displaystyle \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx.
  5. ∫−aaf=2∫0af\displaystyle \int_{-a}^a f = 2\int_0^a f if ff is even, and 00 if ff is odd.
  6. ∫02af=2∫0af\displaystyle \int_0^{2a} f = 2\int_0^a f if f(2a−x)=f(x)f(2a - x) = f(x), and 00 if f(2a−x)=−f(x)f(2a - x) = -f(x).

Exam­ple 5. ∫−22∣x+1∣ dx=∫−2−1(−x−1) dx+∫−12(x+1) dx=12+92=5\displaystyle \int_{-2}^{2}\lvert x + 1 \rvert\,dx = \int_{-2}^{-1}(-x - 1)\,dx + \int_{-1}^{2}(x + 1)\,dx = \tfrac{1}{2} + \tfrac{9}{2} = 5.

The V-shaped graph y = |x + 1| with its corner at x = -1; under it from -2 to 2 lie a small triangle of area 1/2 and a larger triangle of area 9/2, total 5.
Split­ting at the cor­ner x = -1 gives two tri­an­gles: 1/2 + 9/2 = 5.

Exam­ple 6. ∫−11(x5+x3cos⁡x) dx=0\displaystyle \int_{-1}^{1}(x^5 + x^3\cos x)\,dx = 0, with no cal­cu­la­tion at all, because the inte­grand is odd.

Graph of the odd function y = x to the 5th + x cubed cos x from -1 to 1: the shaded part above the axis on the right mirrors the part below on the left, so they cancel.
For an odd inte­grand the two halves can­cel, so the inte­gral is 0.

Exam­ple 7. I=∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle I = \int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx. This looks hard, but prop­erty 4 gives I=∫0π/2cos⁡xcos⁡x+sin⁡x dx\displaystyle I = \int_0^{\pi/2}\frac{\cos x}{\cos x + \sin x}\,dx. Add the two forms and the inte­grand col­lapses to one, so 2I=π2\displaystyle 2I = \tfrac{\pi}{2}, I=π4\displaystyle I = \tfrac{\pi}{4}.

Exam­ple 8. ∫0πxsin⁡x1+cos⁡2x dx\displaystyle \int_0^{\pi}\frac{x\sin x}{1 + \cos^2 x}\,dx. Replace xx by π−x\pi - x and add the two ver­sions: 2I=π∫0πsin⁡x1+cos⁡2xdx=π⋅π2\displaystyle 2I = \pi\int_0^\pi\frac{\sin x}{1 + \cos^2 x}dx = \pi \cdot \tfrac{\pi}{2}, so I=π24\displaystyle I = \tfrac{\pi^2}{4}.

Mixed prac­tice

  1. ∫02(x3+2x) dx\displaystyle \int_0^2(x^3 + 2x)\,dx; ∫1edxx\displaystyle \int_1^e\tfrac{dx}{x}; ∫0π/3cos⁡x dx\displaystyle \int_0^{\pi/3}\cos x\,dx.
  2. ∫01xx2+1 dx\displaystyle \int_0^1\frac{x}{x^2 + 1}\,dx; ∫12dxx(1+log⁡x)2\displaystyle \int_1^2\frac{dx}{x(1 + \log x)^2}.
  3. ∫0π/2cos⁡2x dx\displaystyle \int_0^{\pi/2}\cos^2 x\,dx; ∫01xex dx\displaystyle \int_0^{1}xe^x\,dx.
  4. ∫−33∣x−1∣ dx\displaystyle \int_{-3}^{3}\lvert x - 1 \rvert\,dx.
  5. ∫−π/2π/2sin⁡7x dx\displaystyle \int_{-\pi/2}^{\pi/2}\sin^7 x\,dx; ∫−11(x2+x3) dx\displaystyle \int_{-1}^{1}(x^2 + x^3)\,dx.
  6. ∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx.
  7. ∫2810−xx+10−x dx\displaystyle \int_2^8\frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}\,dx.
  8. ∫0π/2log⁡(sin⁡x) dx\displaystyle \int_0^{\pi/2}\log(\sin x)\,dx (answer: −π2log⁡2\displaystyle -\tfrac{\pi}{2}\log 2; show the key step 2I=∫0π/2log⁡sin⁡2x2 dx\displaystyle 2I = \int_0^{\pi/2}\log\tfrac{\sin 2x}{2}\,dx).
  9. ∫04dxx2+9\displaystyle \int_0^4\frac{dx}{\sqrt{x^2 + 9}}.
  10. ∫13dxx(x+1)\displaystyle \int_1^3\frac{dx}{x(x + 1)}.
  11. ∫01tan⁡−1x dx\displaystyle \int_0^1\tan^{-1}x\,dx.
  12. ∫02[x] dx\displaystyle \int_0^{2}[x]\,dx where [x][x] is the great­est inte­ger func­tion.

Answers to check against

Show answers
  1. 88; 11; 32\displaystyle \tfrac{\sqrt{3}}{2}.
  2. 12log⁡2\displaystyle \tfrac{1}{2}\log 2; with t=1+log⁡xt = 1 + \log x it comes to 1−11+log⁡2\displaystyle 1 - \tfrac{1}{1 + \log 2}.
  3. π4\displaystyle \tfrac{\pi}{4}; 11.
  4. ∫−31(1−x)dx+∫13(x−1)dx=8+2=10\displaystyle \int_{-3}^{1}(1 - x)dx + \int_1^3(x - 1)dx = 8 + 2 = 10.
  5. 00; 23\displaystyle \tfrac{2}{3}.
  6. π4\displaystyle \tfrac{\pi}{4}.
  7. Use prop­erty 3 with a+b=10a + b = 10: 2I=62I = 6, I=3I = 3.
  8. Adding II to its cos⁡\cos ver­sion gives 2I=∫0π/2log⁡sin⁡2x dx−π2log⁡2=I−π2log⁡2\displaystyle 2I = \int_0^{\pi/2}\log\sin 2x\,dx - \tfrac{\pi}{2}\log 2 = I - \tfrac{\pi}{2}\log 2.
  9. log⁡∣x+x2+9∣\log\lvert x + \sqrt{x^2 + 9}\rvert from 00 to 44: log⁡9−log⁡3=log⁡3\log 9 - \log 3 = \log 3.
  10. [log⁡xx+1]13=log⁡34−log⁡12=log⁡32\displaystyle \left[\log\tfrac{x}{x + 1}\right]_1^3 = \log\tfrac{3}{4} - \log\tfrac{1}{2} = \log\tfrac{3}{2}.
  11. π4−12log⁡2\displaystyle \tfrac{\pi}{4} - \tfrac{1}{2}\log 2.
  12. 0+1=10 + 1 = 1.