. Graphical and mechanical computation . e in the neighborhood of B = 10. We therefore tabu-late the differences of /jl in the neighborhood of B = 10. 910111213 A> A2 I33013401320I2501120 10— 20 -30 - 70 -50-60 -130 — 20 — IO Art. 105 GRAPHICAL INTEGRATION 237 For values of B between B = 9 and B = 10, we have^=[io+(2W-i)(-f)+(3«2-6n+2)(-^)] = |(ii-6n-6^). = 0, hence 6 «2 + 6 » — 11 =0, and n = dB For a maximum, dB Therefore, B = B0 + nh = We find the corresponding value of /x by the interpolation formula,ju = 1330 + () (10) + () (-30) + () (-20) = we take
. Graphical and mechanical computation . e in the neighborhood of B = 10. We therefore tabu-late the differences of /jl in the neighborhood of B = 10. 910111213 A> A2 I33013401320I2501120 10— 20 -30 - 70 -50-60 -130 — 20 — IO Art. 105 GRAPHICAL INTEGRATION 237 For values of B between B = 9 and B = 10, we have^=[io+(2W-i)(-f)+(3«2-6n+2)(-^)] = |(ii-6n-6^). = 0, hence 6 «2 + 6 » — 11 =0, and n = dB For a maximum, dB Therefore, B = B0 + nh = We find the corresponding value of /x by the interpolation formula,ju = 1330 + () (10) + () (-30) + () (-20) = we take account of A1 and A2 only, we get ^| = io+(2«- i)(y)=o, or w =| = , and 5= Then fx = 1330 + () (10) + () (-30) = 1337-5- 105. Graphical integration. — Let us find the value of the definiteintegral / f(x) dx or the area under the curve y = f(x) by graphical %)a methods. We draw the curve y = f(x) (Fig. 105a) and along the ordinateat P (x, y) erect the ordinate y whose value is a measure of the area under.
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