write ∫ [4,0] 1/(𝑥−4)^3 𝑑𝑥 as
∫ [4,0] (𝑥−4)^-3 𝑑𝑥
the indefinite integral would be 1/(2(x-4)^2) + c
when we try to evaluate
= (x-4)^-2 / -2 | from 0 to 4
= 1/(2(x-4)^2) | from 0 to 4
= 1/(2(0)..... ahhh, we are dividing by zero
time to look at the graph:
there is a vertical asymptote at x = 4, and x=4 is your upper boundary
You don't have a closed region, so the "area" is infinitely large.
Evaluate the improper integral
∫ [4,0] 1/(𝑥−4)^3 𝑑𝑥
I am stumped please help
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I am having problem with a similar one can you explain it further
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