Evaluate the improper integral
โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ
I am stumped please help
write โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ as
โซ [4,0] (๐ฅโ4)^-3 ๐๐ฅ
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.
I am having problem with a similar one can you explain it further
To evaluate the improper integral โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ, we can follow these steps:
Step 1: Split the integral.
Since the integrand has a singularity at x = 4, we need to split the integral into two parts: one from 0 to slightly below 4 and the other from slightly above 4 to 4.
Let's call the lower part of the integral I1 and the upper part I2.
Step 2: Evaluate I1.
To evaluate I1, we integrate the function 1/(๐ฅโ4)^3 from 0 to a number slightly smaller than 4, such as 3.9.
โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ = โซ [3.9,0] 1/(๐ฅโ4)^3 ๐๐ฅ
Step 3: Integrate I1.
To find the antiderivative of 1/(๐ฅโ4)^3, we can use a substitution. Let u = x - 4, then du = dx.
Plugging this into the integral:
โซ 1/(๐ฅโ4)^3 ๐๐ฅ = โซ 1/u^3 du
= โซ u^(-3) du
= -u^(-2)/2 + C
= -1/(2u^2) + C
Step 4: Evaluate I1.
Now, we can evaluate I1 by plugging in the limits of integration:
I1 = -[ -1/(2(3.9-4)^2)] - [ -1/(2(0-4)^2)]
= -[ -1/(2(-0.1)^2)] - [ -1/(2(-4)^2)]
Step 5: Evaluate I2.
To evaluate I2, we integrate the function 1/(๐ฅโ4)^3 from a number slightly larger than 4, such as 4.1, to 4.
โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ = โซ [4,4.1] 1/(๐ฅโ4)^3 ๐๐ฅ
Step 6: Integrate I2.
Using the same substitution as before with u = x - 4, we can integrate the function:
โซ 1/(๐ฅโ4)^3 ๐๐ฅ = โซ 1/u^3 du
= โซ u^(-3) du
= -u^(-2)/2 + C
= -1/(2u^2) + C
Step 7: Evaluate I2.
Now, we can evaluate I2 by plugging in the limits of integration:
I2 = -[ -1/(2(4.1-4)^2)] - [ -1/(2(4-4)^2)]
= -[ -1/(2(0.1)^2)] - [ -1/(2(0)^2)]
Step 8: Calculate the total integral.
Finally, we can add I1 and I2 to find the value of the total improper integral:
โซ [4,0] 1/(๐ฅโ4)^3 ๐๐ฅ = I1 + I2