Question

Assuming an architecture where the values in its registers are as follows, AR = 80, DR = 105, R6 = 116, R7 = 12 and R8=82 the content of Memory address 80 is 20 and the content of Memory address 85 is 10. Show the content of the 5 registers and the content of the memory addresses 80 and 85 after the following RTL statements.

Answer 1

Answer : I = ∫{x^2 * sin(3*x)} where upper limit is b = +1 and lower limit a=-1.

so the value of I would be calculated by integrating the above expression first with upper limit and then adding the value by putting lower limit.

So, I = ∫{x^2 * sin(3*x)} when b = +1 is putting x = +1

I = 1^2 sin(3*1) = sin3 = 0.05233596

So, I = ∫{x^2 * sin(3*x)} when a = -1 is putting x = -1

I = 1^2 sin(3*-1) = sin(-3) = -0.05233596

Adding both the values we get I = 0.0