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By Trapezium Rule (with 5 segments)x 0 0.2 0.4 0.6 0.8 1y 2 1.8625 1.71830 1.56981 1.41994 1.27165≈1⁄2*(0.2) [2+1.27165+ 2( 1.8625 + 1.71830 + 1.56981 + 1.41994)]≈1⁄10 *16.41283 = 1.641By Substitution/ Pa...
(2+kx)-3 = (2-3)(1+kx/2)-3 = (2-3)(1+(-3)(kx/2) + [(-3)(-4)]/2! (kx/2)2 + [(-3)(-4)(-5)]/3! (kx/2)3 +... )= 1⁄8 [1 -(3kx/...
If we were to expand the brackets for this question we might be able to get to an answer however there is a much simpler way of solving the problem. As this question is made up of two expressions of x mul...
Taking upwards to be positive, and using the 'suvat' equation s=ut+1/2at2 we know that u=u, a=-g and s=0 when the particle returns to the ground. Then we solve for t:0=ut-1/2gt2JVAnswered by Jackie V. • Maths tutor3089 Views
First use the substitution to find du/dx which is 2x. From this we now know that dx= du/2x (just re-arranging.) Substituting that into the integral we now get (6x3+4x)/ (sqrt (u) x 2x) du. Canc...
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