Question:

Find the area under the curve defined by the function f(x) = x^2 + 3x + 2 on the interval [1, 5].

Step-by-step Solution:

To find the area under the curve, we need to integrate the given function over the given interval.

1. First, let's integrate the function f(x) = x^2 + 3x + 2. ∫(x^2 + 3x + 2) dx = ∫x^2 dx + ∫3x dx + ∫2 dx

2. Applying the power rule of integration, we have: = (1/3)x^3 + (3/2)x^2 + 2x + C

3. Now, we need to evaluate this antiderivative over the interval [1, 5]. Substitute x = 5 into the antiderivative: = (1/3)(5)^3 + (3/2)(5)^2 + 2(5) + C = (1/3)(125) + (3/2)(25) + 10 + C = 125/3 + 75/2 + 10 + C

   Substitute x = 1 into the antiderivative: = (1/3)(1)^3 + (3/2)(1)^2 + 2(1) + C = (1/3)(1) + (3/2)(1) + 2 + C = 1/3 + 3/2 + 2 + C

4. Now subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit to find the definite integral: = [(125/3) + (75/2) + 10 + C] - [(1/3) + (3/2) + 2 + C] = 125/3 + 75/2 + 10 - 1/3 - 3/2 - 2 - C

   Simplifying the expression: = 41.833 - 3.167 - C = 38.666 - C

5. The area under the curve defined by the function f(x) = x^2 + 3x + 2 on the interval [1, 5] is approximately 38.666 - C square units.

Note: The constant of integration, C, cancels out when subtracting the antiderivative, so it does not affect the final result.