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Evaluate the integral.

π
∫ f(x) dx where f(x) = { 2 sin x if 0 < x < π/2
⁰ = { 3 cos x if π/2 ≤ X ≤ π
What is wrong with the equation?
₁ ₁
∫ x⁻³ dx = x⁻²/-2] = - 4/9
⁻³ ⁻³
a. There is nothing wrong with the equation.
b. f(x) = x⁻³ is continuous on the interval [-3, 1] so FTC2 cannot be applied.
c. The lower limit is less than 0, so FTC2 cannot be applied.
d. f(x) = x⁻³ is not continuous on the interval [-3, 1] so FTC2 cannot be applied.
e. f(x) = x⁻³ is not continuous at x = -3, so FTC2 cannot be applied.

1 Answer

2 votes

Final answer:

The piecewise integral must be evaluated in two parts due to its definition. The expression for the separate integral with the function x^(-3) is incorrect because the function is not continuous on the interval [-3, 1], which precludes the application of the second part of the Fundamental Theorem of Calculus.

Step-by-step explanation:

You are given two pieces of information in your math problem: a definite integral of a piecewise function and a separate integral calculation that you have concerns about. First, let's address the piecewise function integral. To evaluate the integral π∫ f(x) dx where f(x) is defined as 2 sin x for 0 < x < π/2 and 3 cos x for π/2 ≤ x ≤ π, you need to split the integral into two at x = π/2 and evaluate each integral separately, then sum the results.

For the second integral ∫ x⁻³ dx from -3 to 1, there is an issue. The correct answer is (d): f(x) = x⁻³ is not continuous on the interval [-3, 1] so FTC2 (the second part of the Fundamental Theorem of Calculus) cannot be applied. Specifically, the function x⁻³ is not defined at x = 0, which makes the function discontinuous on the interval, hence the problem with the integral equation given. The function must be continuous on the entire interval for FTC2 to be applicable.

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