Determine whether the integral is divergent or convergent. If it is convergent, evaluate it and enter that value as your answer. If it diverges to infinity, state your answer as "INF" (without the quotation marks). 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".
The integral from 0 to 7 of -1/((x-5)^2)dx
I keep getting 0.7 but apparently I am wrong.
Apparently it diverges to negative infinity. Can someone please explain to me why?
I do not know how you got 0.7, negative infinity seems to me correct because there is an asymptote at x=5.
For improper integrals where the vertical asymptote is between the integration limites, we have to be careful not just evaluate the integral at the two limits.
Instead, we will subdivide the integral into two, from 0 to 5 and from 5 to 7.
If both integrals converge, then you have convergence and the integral is defined.
If there are more vertical asymptotes, you will need to subdivide accordingly.
To determine whether the integral is convergent or divergent, we need to check if it meets the conditions of the Integral Test for convergence.
For the given integral from 0 to 7 of -1/((x-5)^2)dx, let's check the behavior of the integrand.
The integrand, -1/((x-5)^2), has a singularity at x = 5. When x approaches 5, the denominator approaches 0 and the function approaches negative infinity.
We can split the integral into two parts: one from 0 to 5 and another from 5 to 7, as the integrand has a singularity at x = 5.
For the integral from 0 to 5, the integrand is negative and becomes infinitely large as x approaches 5. This indicates that the integral diverges to negative infinity (MINF).
For the integral from 5 to 7, the integrand is negative but remains bounded as x approaches 5. Therefore, the integral is finite.
Putting these together, the overall behavior of the integral is that it diverges without being infinity or negative infinity (DIV).
Hence, the integral from 0 to 7 of -1/((x-5)^2)dx is divergent (DIV).
To determine whether the integral is divergent or convergent, we need to examine the behavior of the integrand as x approaches the boundaries of the interval of integration.
In this case, the integrand is -1/((x-5)^2), which can be rewritten as (-1)/(x-5)^2.
Let's first look at the behavior of the integrand as x approaches 5 from below. As x gets closer and closer to 5 from the left-hand side (x < 5), the denominator (x-5)^2 becomes smaller and smaller, leading to a larger and larger negative value for the integrand. As a result, the integrand approaches negative infinity as x approaches 5 from below.
Next, let's examine the behavior of the integrand as x approaches 5 from above. As x gets closer and closer to 5 from the right-hand side (x > 5), the denominator (x-5)^2 becomes smaller and smaller, but now it is always positive since x is greater than 5. As a result, the integrand also approaches negative infinity as x approaches 5 from above.
Since the integrand approaches negative infinity as x approaches both 5 from below and from above, the integral diverges. Therefore, the answer is "DIV". It is not appropriate to evaluate the integral as a numeric value because it is not convergent.