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Nat2105 [25]
2 years ago
14

What is -6x - 10(x - 4)​

Mathematics
1 answer:
weeeeeb [17]2 years ago
4 0
-16x+40 is the solution of this problem.
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Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F
Svet_ta [14]

Answer:

A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36

Step-by-step explanation:

The equations are:

y = x^{2} + 2x + 3

y = 2x + 12

The two graphs intersect when:

x^{2} + 2x + 3 = 2x + 12

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x_{1}  = 3\\x_{2}  = -3

To find the area under the curve for the first equation:

A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx

To find the area under the curve for the second equation:

A_{2} = \int\limits^3__-3}{2x + 12} \, dx

To find the total area:

A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx

Simplifying the equation:

A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).  

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