Which of the following limits is equal to ∫31sin(x3+2)ⅆx ?
limn→∞∑k=1nsin((1+2kn)3+2)2n
We can start by using the substitution method to find the indefinite integral of sin(x³ + 2). Let u = x³ + 2, then du/dx = 3x² and dx = du/3x². Plugging these into the original integral, we have:
∫3¹ sin(x³ + 2)ⅆx = ∫u₁¹ sin(u) * (1/3x²) du = (1/3)∫u₁¹ sin(u)/x² du
Using u-substitution and the fact that the integral of sin(u) is -cos(u), we get:
= (1/3)[-cos(u)]u=3¹
= (1/3)[-cos(3³ + 2) + cos(2)]
= (1/3)[cos(2) – cos(29)]
Therefore, the limit is equal to (1/3) [cos(2) – cos(29)].
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