Which of the following limits is equal to ∫31sin(x3+2)ⅆx ?
limn→∞∑k=1nsin((1+2kn)3+2)2n
To solve this problem, we need to use the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) – F(a).
Let’s start by finding an antiderivative of sin(x^3 + 2). Since the derivative of x^3 is 3x^2, we can use a u-substitution to simplify the integral:
Let u = x^3 + 2, then du/dx = 3x^2, and dx = du/3x^2
∫sin(x^3+2)dx = (1/3)∫sin(u)du
We know that the antiderivative of sin(u) is -cos(u), so our integral now becomes:
(1/3)[-cos(x^3 + 2)] + C
where C is the constant of integration.
Using the fundamental theorem of calculus, we can evaluate the definite integral ∫3^1sin(x^3+2)dx by subtracting the value of the antiderivative at x=1 from its value at x=3:
∫3^1sin(x^3+2)dx = [-cos(1^3 + 2)]/3 – [-cos(3^3 + 2)]/3
= [-cos(3) + cos(29)]/3
Therefore, the limit that is equal to the integral ∫3^1sin(x^3+2)dx is [-cos(3) + cos(29)]/3.
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