Can someone solve this integral for me? My calculator just gives me a decimal approximation but I need an exact number. Thanks! :)
pi[{3ln(3)^2 - 6ln (3) + 6} - {ln(1)^2 - 2 ln(1) + 2}]
That does not look like a function to be integrated. It looks like a number to be calculated. You can start out simplifying it by recognizing that
ln(1) = 0.
When you write ln(3)^2, do you mean
[ln(3)]^2 or ln[3^2]? They are not the same. I will assume the latter.
ln [3^2] = 2 ln 3, so
3 ln[(3)^2] - 6ln (3) = 0
That leaves you with pi (6 - 4) = 2 pi
I'm confused. When I wrote it out by hand, I got:
ð(3ln(9)-6ln(3)+6)-(ln(1)-2ln(1)+2)
ð(ln(93)-ln(36)+6)- 2 (since ln(1)=0)
ð(ln(729)-ln(729)+6)-2
6ð-2
the weird symbol is pi
and that 93 is 9^3
and that 36 is 3^6
To solve this integral, we will need to work through it step by step. Given the expression you provided:
∫[π(3ln(3)^2 - 6ln(3) + 6 - ln(1)^2 + 2ln(1) - 2)]
Let's simplify and rewrite it before integrating:
∫[π(3ln(3)^2 - ln(1)^2 - 6ln(3) + 2ln(1) + 4)]
Now, since ln(1) equals 0, we can simplify further:
∫[π(3ln(3)^2 - 6ln(3) + 4)]
To integrate this expression, we will use the power rule for integrals. The general form of the power rule is:
∫[x^n] dx = (1/(n+1)) * x^(n+1) + C
Applying the power rule to our integral:
∫[3ln(3)^2 - 6ln(3) + 4] = (π/3) * ln(3)^3 - (π/2) * ln(3)^2 + 4π * x + C
So, the exact value of the integral is:
(π/3) * ln(3)^3 - (π/2) * ln(3)^2 + 4π * x + C
Please note that "x" is the variable of integration, so if you wanted a definite integral within a specific range, you would need to substitute the upper and lower limits of integration into the expression and evaluate it accordingly.