computing definite integral by appealing to geometric formulas Graph the following definite integral then find the exact area of the shaded region justify your answer for each problem

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.