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Evaluate the integral by interpreting it in terms of areas: .
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Step 1: Recognize that the integral \( \int_0^9 (3x - 2) \, dx \) represents the net area between the curve \( y = 3x - 2 \) and the x-axis over the interval \( [0, 9] \). This involves finding the areas of geometric shapes formed by the curve and the x-axis.
Step 2: Determine where the curve \( y = 3x - 2 \) intersects the x-axis by solving \( 3x - 2 = 0 \). This gives \( x = \frac{2}{3} \). The curve changes from being below the x-axis to above the x-axis at this point.
Step 3: Split the integral into two parts: \( \int_0^{\frac{2}{3}} (3x - 2) \, dx \) and \( \int_{\frac{2}{3}}^9 (3x - 2) \, dx \). The first part corresponds to the area below the x-axis (negative contribution), and the second part corresponds to the area above the x-axis (positive contribution).
Step 4: Compute the areas geometrically. The curve \( y = 3x - 2 \) forms a triangle below the x-axis from \( x = 0 \) to \( x = \frac{2}{3} \), and a trapezoid above the x-axis from \( x = \frac{2}{3} \) to \( x = 9 \). Use the formulas for the area of a triangle and trapezoid to calculate these areas.
Step 5: Add the absolute value of the negative area (triangle) to the positive area (trapezoid) to find the total net area. This sum corresponds to the value of the integral \( \int_0^9 (3x - 2) \, dx \).