The answer is 2/243(82^1.5 - 1) or about 6.1... but I don't know how to get to that (most especially the 2/243)... I know I can send 2/3 out of the integral but sometime or another but I'll have to send out 82 as well to get 2/243 and I don't see where I can do that.
Anonymous
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Yes, you use substitution method and then, FTC (First Fundamental Theorem of Calculus) to solve that problem.
∫0 to 1 (1 + 81x)^(½) dx
Let u = 1 + 81x and...
du/dx = 81
du = 81 dx
du/81 = dx
Note that you can substitute the limits for u function.
u(0) = 1
u(1) = 82
This yields:
==> 1/81 ∫1 to 82 (u)^(½) du
==> 1/81*(2/3)*u^(3/2) | 1 to 82
==> 2(u)^(3/2)/243 | 1 to 82
Finally, by FTC:
==> (2(82)^(½) - 2(1)^(½))/243
==> (2(82)^(0.5) - 2)/243
or
(2/243) * (82^(1.5) - 1)
I hope this helps!
taisho_tentei
U substitution.
let u = 1 + 81 x
du = 81 dx
1/81 â« u^.5 du //Reverse power rule
2/243 (1 + 81x) ^ (3/2).
//Fundamental theorem f(1) - f(0) should give you your answer.
?
u = 1+81x
du = 81dx
(1/81)â«u^0.5 du
= (2/3)(1/81)u^1.5
= (2/243)(u^1.5)
= (2/243)(1+81x)^1.5
= 6.111 - 0.008
= 6.103