Answer :
the total area of the region between the x-axis and the graph of y = x³ + 6x² + 8x over the interval -4 ≤ x ≤ 0 is -16 square units.
To find the total area of the region between the x-axis and the graph of y = x³ + 6x² + 8x over the interval -4 ≤ x ≤ 0, we need to evaluate the definite integral of the absolute value of the function over that interval.
Since the graph is above the x-axis for the given interval, the total area can be found by integrating the function y = x³ + 6x² + 8x from -4 to 0.
Total area = ∫(from -4 to 0) |x³ + 6x² + 8x| dx
Let's find the integral:
∫(from -4 to 0) |x³ + 6x² + 8x| dx = ∫(from -4 to 0) (x³ + 6x² + 8x) dx
Now, integrate term by term:
∫(x³) dx = (1/4) x^4
∫(6x²) dx = 2x³
∫(8x) dx = 4x²
Evaluate the definite integral from -4 to 0:
Total area = [(1/4) (0)^4 + 2(0)^3 + 4(0)²] - [(1/4) (-4)^4 + 2(-4)^3 + 4(-4)²]
Total area = [0 + 0 + 0] - [16 + (-32) + 64]
Total area = 0 - 16 + 32 - 64
Total area = -16
So, the total area of the region between the x-axis and the graph of y = x³ + 6x² + 8x over the interval -4 ≤ x ≤ 0 is -16 square units.
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