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
Read more about areas at:
#SPJ1
You might be interested in
Answer:
The value of k is 5.
Step-by-step explanation:
Since we know we can convert function's quadratic form to vertex form by completing the square.
f(x)=x2?6x+14
We will add and subtract 9 to our equation in order to complete the square.
Upon completing the square and combining like terms we will get,
Upon comparing Willie's vertex form with our expression we can see that k is 5.
Therefore, the value of k is 5.