Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV.
�ç5 to −�‡ (1/(x^2+1))dx
If you mean
∫[5,-∞] 1/(x^2+1) dx
then you have
arctan(x) [5,-∞]
= -π/2 - arctan(5)
To determine whether the given integral is divergent or convergent, we can apply the integral test. The integral test states that if a function f(x) is positive, continuous, and decreasing on the interval [a, ∞), then the improper integral ∫(a to ∞) f(x) dx is convergent if and only if the corresponding series Σ (n = a to ∞) f(n) is convergent.
In this case, the function f(x) = 1/(x^2+1) is positive and continuous in the interval (-∞, ∞). However, it's not decreasing. So, we cannot directly apply the integral test.
Instead, we can apply a different test called the limit comparison test. The limit comparison test states that if f(x) and g(x) are positive functions on [a, ∞), and if lim (x→∞) f(x)/g(x) = L, where L is finite and positive, then both ∫ (a to ∞) f(x) dx and ∫ (a to ∞) g(x) dx either both converge or both diverge.
In this case, let's compare the given function with g(x) = 1/x^2. Both f(x) = 1/(x^2+1) and g(x) = 1/x^2 are positive functions on [5, ∞), and as x approaches ∞, the limit of f(x)/g(x) can be calculated:
lim (x→∞) (1/(x^2+1))/(1/x^2)
= lim (x→∞) x^2/(x^2+1)
= lim (x→∞) 1/(1+1/x^2)
= 1
Since the limit is finite and positive, we can conclude that either both ∫ (5 to ∞) f(x) dx and ∫ (5 to ∞) g(x) dx converge or both diverge.
The integral ∫ (5 to ∞) g(x) dx = ∫ (5 to ∞) 1/x^2 dx is a known convergent integral that can be evaluated using basic integration techniques:
∫ (5 to ∞) 1/x^2 dx = [ -1/x ] (from 5 to ∞)
= [ -1/∞ ] - [ -1/5 ]
= 0 + 1/5
= 1/5
Therefore, since the integral ∫ (5 to ∞) g(x) dx converges to a finite value, we can conclude that the original integral ∫ (5 to ∞) f(x) dx also converges, and its value is 1/5.