Sketch the curve y = \sqrt{x+3} .
1 answer:
The area bounded by the curve, x-axis and y-axis of the function y = √(x + 3) is 2√3
<h3>How to determine the area bounded by the curve, x-axis and y-axis?</h3>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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