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Area under the curve of x^3 - 3x^2 + 2x + 1

Created by @nathanedwards
 at November 23rd 2023, 10:36:42 pm.
AP_Calculus_AB-Questions
#area-under-curves
#definite-integral
#ap-calculus-ab

Sure, here's an AP Calculus AB exam question about area under curves:

Question: Let f(x) = x^3 - 3x^2 + 2x + 1. Find the area of the region bounded by the curve y = f(x), the x-axis, and the lines x = 0 and x = 2.

Answer: To find the area of the region bounded by the curve y = f(x), the x-axis, and the lines x = 0 and x = 2, we need to calculate the definite integral of f(x) from x = 0 to x = 2.

The definite integral of f(x) from a to b can be calculated using the following formula:

∫abf(x) dx \int_{a}^{b} f(x) \, dx

First, we need to find the integral of f(x):

∫f(x) dx=∫(x3−3x2+2x+1) dx \int f(x) \, dx = \int (x^3 - 3x^2 + 2x + 1) \, dx

Using the power rule for integration, we have:

∫x3 dx−∫3x2 dx+∫2x dx+∫1 dx \int x^3 \, dx - \int 3x^2 \, dx + \int 2x \, dx + \int 1 \, dx
=14x4−x3+x2+x+C = \frac{1}{4}x^4 - x^3 + x^2 + x + C

Now, to find the area of the region bounded by the curve y = f(x), the x-axis, and the lines x = 0 and x = 2, we evaluate the definite integral of f(x) from x = 0 to x = 2:

∫02f(x) dx=[14x4−x3+x2+x]02 \int_{0}^{2} f(x) \, dx = \left[ \frac{1}{4}x^4 - x^3 + x^2 + x \right]_{0}^{2}
=(14(2)4−(2)3+(2)2+2)−(14(0)4−(0)3+(0)2+0) = \left( \frac{1}{4}(2)^4 - (2)^3 + (2)^2 + 2 \right) - \left( \frac{1}{4}(0)^4 - (0)^3 + (0)^2 + 0 \right)
=(14(16)−8+4+2)−0 = \left( \frac{1}{4}(16) - 8 + 4 + 2 \right) - 0
=4−8+4+2 = 4 - 8 + 4 + 2
=2 = 2

Therefore, the area of the region bounded by the curve y = f(x), the x-axis, and the lines x = 0 and x = 2 is 2 square units.

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