Find the area of the region under the graph of the function f on the interval [-27, -1].
f(x)= 2-3square root x
? square units
To find the area under the graph of the function f(x) = 2 - 3√x on the interval [-27, -1], we can use definite integration.
Step 1: Start by finding the antiderivative of f(x).
The derivative of √x is (1/2)(1/√x), so the derivative of -3√x is -3(1/2)(1/√x) = -3/(2√x).
Integrating -3/(2√x) gives:
∫ (-3/(2√x)) dx = -3/2 ∫ (1/√x) dx
Using the power rule for integration, the antiderivative of (1/√x) is 2√x:
= -3/2 * 2√x + C, where C is the constant of integration.
Simplifying, we have:
= -3√x + C
Step 2: Evaluate the definite integral using the antiderivative.
To find the area under the graph of f(x) on the interval [-27, -1], we need to evaluate the definite integral:
∫[-27, -1] (-3√x) dx
Using the antiderivative we found earlier, we have:
= [-3√x] evaluated from -27 to -1
= -3√(-1) - (-3√(-27))
= -3√(-1) + 3√(-27)
Since the square root of a negative number is not a real number, we need to use imaginary numbers to simplify the result.
√(-1) can be written as i, where i is the imaginary unit (√(-1) = i).
Similarly, √(-27) can be split into √((-1)(3^3)) = 3√(-1) = 3i.
Therefore, the result becomes:
= -3i + 3(3i)
= -3i + 9i
= 6i
So, the area under the graph of the function f(x) = 2 - 3√x on the interval [-27, -1] is 6i square units.
Note: Since the function involves square roots of negative numbers, the result is in the form of an imaginary number.