Question

CAN SOMEONE HELP WITH THIS QUESTION?

Answer 1

39 1/3 = 118/3 square units

Explanation:

You want the area under the curve y = -x² +9x on the interval [3, 5].

Integral

The fundamental theorem of calculus tells us the integral of the function will be its anti-derivative. That is, for f(x) = -x² +9x, we want the function F(x) whose derivative is f(x). The reverse of the power rule for derivatives can be used:

`F(x)=-(x^3)/(3)+9(x^2)/(2)`

Area

The area of interest is ...

F(5) -F(3) = -1/3(5³ -3³) +9/2(5² -3²) = -98/3 +72

F(5) -F(3) = 39 1/3

The area under the curve between x=3 and x=5 is 39 1/3 square units.