Please help me answer this homework question.---**Question Stem:**
A curve is defined by the equation: y = 3x^2 - 2x + 1
(a) Find the area enclosed between the curve and the x-axis between the limits x = 1 and x = 4.
**Other Relevant Text:**
Please help me answer this homework question.
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We need to find the area between the curve y equals 3x squared minus 2x plus 1 and the x-axis, from x equals 1 to x equals 4. This is a definite integral problem where we'll integrate the function over the given interval.
To find the area, we set up the definite integral from 1 to 4 of the function 3x squared minus 2x plus 1. First, we need to find the antiderivative. The antiderivative of 3x squared is x cubed, the antiderivative of negative 2x is negative x squared, and the antiderivative of 1 is x. So our antiderivative F of x equals x cubed minus x squared plus x.
Now we evaluate the antiderivative at both limits. At x equals 4: F of 4 equals 4 cubed minus 4 squared plus 4, which is 64 minus 16 plus 4, equals 52. At x equals 1: F of 1 equals 1 cubed minus 1 squared plus 1, which is 1 minus 1 plus 1, equals 1. These points on the curve show where we're evaluating our antiderivative.
Finally, we apply the Fundamental Theorem of Calculus. The area equals F of 4 minus F of 1, which is 52 minus 1, equals 51 square units. This shaded region represents the exact area under the curve y equals 3x squared minus 2x plus 1 between x equals 1 and x equals 4. The definite integral has given us the precise answer of 51 square units.
To summarize our solution: We found the area between the curve y equals 3x squared minus 2x plus 1 and the x-axis from x equals 1 to x equals 4. We set up the definite integral, found the antiderivative x cubed minus x squared plus x, evaluated it at both limits, and calculated F of 4 minus F of 1 equals 52 minus 1 equals 51. Therefore, the area is 51 square units.