Basically for these type of questions, the shaded area can be represented by integrals. It’s important to note that a definite integral between two integration limits represent the area under the graph between the boundaries.

Where the curve lies above the x-axis, this area under the graph means the region below the curve to the x-axis.

So for example in 1b, the area under the graph is represented by the integral of

1 + sin x

from x = 0 to x = 3pi/2.

The final value obtained in this definite integral will be your value for the area under the graph.

Of course, there will be complications in the area to be calculated at a later stage of this chapter.

for qn1 i managed to do part c, for b i subbed in 3pi/2 and 0 into the integral of 1+sinx but it was wrong
for d, i got 6.90 instead of 6.91 so im not sure if that is fine...

is the modulus only used when the shaded region is below the xaxis and at the left of yaxis?

Integrating 1 + sin x gives

x - cos x for 0 to 3pi/2
= (3pi/2 - cos 3pi/2) - (0 - cos 0)
= 3pi/2 - 0 - (0 - 1)
= 3pi/2 + 1

As for the 6.90 and 6.91 thing, it should just be a rounding error.

Modulus is applied on those regions because the integrals do not recognise the meaning of the calculated integral at all. In other words, the integral is only capable of spewing out a value without knowing the context of calculating areas which cannot be negative.

for 2b, i integrated 4sin0.5x-cos2x to - 8cos0.5x-1/2sin2x, subbed 0 and - 1.05 in and modulus, i got the answer of 0.01865.... which is wrong. if there anything wrong with the integration?

Try subbing in the numbers again, or check that you have input the values in radian mode. Remember that cos 0 = 1 and not 0.

thx :) subbed the numbers again and got the correct answer

there is this qn :
sketch the curves y=sinx and y=cosx for 0<= x <= pi/2. calculate the area of the shaded region enclosed by
a) the curves and the x-axis
b) the curves and the y-axis

the answer for a is (2-sqrt2) and b is (sqrt2-1) but i cannot seem to get it tho, got 1units^2 for a instead