First integral
Since [tex]f(x)=3x+1[/tex] is a linear function, its graph is a line. This means that you can decompose the area under its graph as the sum of a right triangle and a rectangle, as shown in the attached image.
All the dimensions are fairly easy to deduce: we have
[tex]AC = 2,\ CB=6[/tex]
So, the area of the triangle is
[tex]A_t = \dfrac{2\times 6}{2}=6[/tex]
For the rectangle, we have
[tex]A_r = AC\times CE = 2\times 4=8[/tex]
So, the total area is [tex]6+8=14[/tex]
Second integral
We have
[tex]y = \sqrt{1-x^2} \implies y^2=1-x^2 \implies x^2+y^2=1[/tex]
So, this function is the upper half of the unit circle (the two halves are the functions [tex]y=\pm\sqrt{1-x^2}[/tex])
Since the area of the unit circle is [tex]\pi[/tex], the value of the integral is [tex]\frac{\pi}{2}[/tex]
Third integral
It follows the exact same logic as the first one, you only have to adjust the numbers.
