Question

Sketch the curve y = \sqrt{x+3} .

Answer 1

The area bounded by the curve, x-axis and y-axis of the function y = √(x + 3) is 2√3

How to determine the area bounded by the curve, x-axis and y-axis?

The curve is given as:

y = √(x + 3)

The area bounded by the curve, x-axis and y-axis is when x = 0 and y = 0

When y = 0, we have:

0 = √(x + 3)

This gives

x = -3

So, we set up the following integral

A = ∫ f(x) d(x) (Interval a to b)

This gives

A = ∫ √(x + 3) d(x) (Interval -3 to 0)

When the above is integrated, we have:

A = 1/3 * [2(x + 3)^(3/2)] (Interval -3 to 0)

Expand

A = 1/3 * [2(0 + 3)^3/2 - 2(-3 + 3)^3/2]

This gives

A = 1/3 * 2(3)^3/2

Apply the law of indices

A = 2(3)^1/2

Rewrite as:

A = 2√3 or 3.46

Hence, the area bounded by the curve, x-axis and y-axis of the function y = √(x + 3) is 2√3

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