Question

Integral Calculus D:

Answer 1

\[\large\rm A(\color{orangered}{x})=\int\limits_{-2}^{\color{orangered}{x}}f(t)dt\qquad\qquad\qquad\qquad F(\color{orangered}{x})=\int\limits_4^{\color{orangered}{x}}f(t)dt\]

Answer 2

a. A(-2)\[\large\rm A(\color{orangered}{-2})=\int\limits\limits_{-2}^{\color{orangered}{-2}}f(t)dt\]See how the x value doesn't change?
We won't have a "base" for our area. This is always zero.\[\large\rm A(-2)=0\]

Answer 3

b. F(8)\[\large\rm F(\color{orangered}{8})=\int\limits\limits_4^{\color{orangered}{8}}f(t)dt\]This is defined to be the area under the graph f(t),
from t=4 to t=8.

In the illustration provided, that corresponds to the orange area.
The value got cut off when you shared it though,
so whatever that line pointing to the orange area says.
\[\large\rm F(8)=orange~area\]

Answer 4

c. A(4)\[\large\rm A(\color{orangered}{4})=\int\limits\limits_{-2}^{\color{orangered}{4}}f(t)dt\]
Notice that the blue area is the integral of f(t),
from t=-2 to t=0,
and the green area is the integral of f(t),
from t=0 to t=4.

So together they give us the area under f(t),
from t=-2 to t=4.\[\large\rm A(4) = blue~area~+~green~area\]

Answer 5

Think you can do the rest?
Math is fun, yayyyy :D