How do you integrate (1+81x)^0.5 from 0 to 1?

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!

U substitution.

let u = 1 + 81 x

du = 81 dx

1/81 ∫ 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)∫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