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R is the first quadrant region enclosed by the x-axis and the curve y = 2x + b, and the line x = b, where b > 0. Find the value of b so that the area of the region R is 288 square units

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LammettHash

The area of \(R\) is given by the integral

\(\displaystyle\int_0^b(2x+b)\,\mathrm dx=288\)

Integrating yields

\((x^2+bx)\bigg|_{x=0}^{x=b}=(b^2+b^2)-(0+0)=2b^2\)

So you have

\(2b^2=288\implies b^2=144\impiles b=12\)

R is the first quadrant region enclosed by the x-axis and the curve y = 2x + b, and the line x = b, where b > 0. The value of b is 12 unit.

How do find the surface area of some object?

Find the area that its outer surfaces possess. The Sum of all those surfaces areas is the surface area of the considered object.

R is the first quadrant region enclosed by the x-axis and the curve y = 2x + b, and the line x = b, where b > 0.

The area is given by the integral

\(A = \int\limits^{b}_0(2x + b )\, dx = 288\\\\= (x^{2} + bx) |^b_0\\\\= 2b^2\\\\\)

Integrating yields

\(2b^2 = 288\\\\b = \sqrt{144} \\\\b = 12\)

Learn more about the surface area here:

https://brainly.com/question/15096475

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