Using the substitution of u=6x+5 find the value of the area under the curve f(x)=(2x-3)(6x+%)^1/2 bounded between x=1 and x=1/2 to 4 decimal places.
dx=du/6 => (u-5)/6=x So the integral is now (2((u-5)/6)-3)(u^1/2) du/6 Which through simplifying becomes (1/36)(2u-28)(u^1/2)du = (1/36)(2u^3/2 -28u^1/2)du After integrating becomes (1/36)(4(u^5/2)/5 -56(u^3/2)/3) Bounded between u=11 and u=8 by the substitution After evaluating we reach our final answer of -2.2889 to 4dp
JT
Related Maths A Level answers
All answers ▸The arithmetic series is given by (k+1)+(2k+3)+(3k+5)+...+303. a)Find the number of term in the series in terms of k. b) Show that the sum of the series is given by (152k+46208)/(k+2). c)Given that S=2568, find k.
A uniform ladder is leaning against a smooth wall on a rough ground. The ladder has a mass of 10 kilograms and is 4 metres long. If the ladder is in equilibrium, state an equation for the coefficient of friction of the ground
(A) express 4^x in terms of y given that 2^x = y. (B) solve 8(4^x ) – 9(2^x ) + 1 = 0
What is 'differentiation'?