Wednesday

April 16, 2014

April 16, 2014

Posted by **maths** on Monday, March 24, 2008 at 2:59pm.

(i) f (t) = 6 cos(3t) + 5e^−10t

(ii) g(x) = 21-12x^3/ x (x > 0)

(iii) h(u) = cos^2( 1/8 u)

(b) Evaluate: (this big F sign at the start, 5 at the top and 1 at the bottom)

5 1/4x (7 + 6x^2) dx

F

1

(c)

(i) Write down a deﬁnite integral that will give the value of the area under the curve y = x^2 cos(2x) between x = 3/4 pie and x = pie.

(You are not asked to evaluate the integral by hand.)

(ii) ﬁnd the area described in part (c)(i), giving your

answer correct to three decimal places

MY ANSWERS:

for a) ii)

6cos (3t) dx = -4sin (3t) + c

5e^10t dx = 1/5 e^-10t + c

ii) g (x) = 21- 12x^3 / x (x>0)

diverse through by x

g (x) = 21/ x - 12x^3

21/ x - 12x^3 dx = 12Ln x - x^4 + c

iii) h (u) = cos^2 (1/8u)

using the second version of the double angled formula for cos (2delta)

=1/2 cos (2x 1/8 u) + 1/2 du

= 1/2 / 28 sin (2/8u) + 1/2 x + c

= 6/4 sin (2/8u) + 1/2u + c

i dont know whether thats right, and i dont know how to do the rest.

could some body please help me out. thanks!

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