Question

Calculate the area, in square units, bounded by g(x)=4x+8 and f(x)=x2+19x+62 over the interval [−9,−6].

Answer 1

Try this solution (see the attachmed pictures), modify the design according to the local requirements.

Note, the points (-6;-16) and (-9;-28) are the intersection points of the parabola and the line, given in the condition.

Answer 2

Using integrals, the area bounded by the two curves in the interval is of 415 units squared.

How is the area under a graph found?

The area under the given graph of a curve bounded above by a function g(x) and below by a function f(x) over an interval [a,b] is given by:

$A = \int\limits^a_b f(x) - g(x) \, dx$

We have given functions are

g(x)=4x+8

f (x) = x^2+19x+62

The interval is [−9,−6].,

Then:

$A = \int\limits^a_b f(x) - g(x) \, dx\\\\A = \int\limits^{-6}_{-9} x^2+19x+62 - 4x-8 \, dx\\\\A = \int\limits^{-6}_{-9} x^2+15x+ 54 \, dx\\\\A = x^3 / 2+15x^2 /2+ 54x|^{-6}_{-9} \, dx$

Applying the Fundamental Theorem of Calculus,

We get

$A = -6^3 / 2+15(-6)^2 /2+ 54(-6) - (-9)^3 / 2+15(-9)^2 /2+ 54(-9)$

A = 415

The area is of 415 units squared.

To learn more about integrals ;

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