Calculus antiderivatives using u-substitution?
I don't understand u-substitution in calculus. I'm supposed to evaluate the definite integral over a certain interval. Could somebody please explain how to solve this problem? I don't care about the answer, I just need to understand how to solve the problem! Thank you!
Problem: solve on an interval of 0-4
*integral* (*square root*(2X+1))dx
- RyomaLv 79 years agoBest Answer
U-substitution is the inverse of the chain rule. However, as it is an inverse, instead of multiplying the derivative of the inner function, it is as if you divide by it. So the trick is to find a function and also find its derivative being multiplied so you can go from ∫ f(g(x)) * g'(x) dx to ∫ f(u) du (notice that if you apply the chain rule to F(u), the integral of f(u), you get f(u) * u' = f(g(x)) * g'(x)).
∫ √(2x + 1) dx
Notice that we can integrate √x = x^(1/2) using the power rule, and that the function we have is of a similar form. We see that the derivative of 2x + 1 is 2, which we don't have in the integrand. Luckily though, we can multiplying/dividing by 2:
∫ √(2x + 1) dx
= ∫ 2/2 * √(2x + 1) dx
= 1/2 * ∫ √(2x + 1) * 2 dx
Now, we make make the substitution u = 2x + 1, du = 2 dx:
1/2 * ∫ √(2x + 1) * 2 dx
= 1/2 * ∫ √u du
= 1/2 * ∫ u^(1/2) du
= 1/2 * [u^(1/2 + 1)/(1/2 + 1)]
= 1/2 * [u^(3/2)/(3/2)]
= 1/3 * u^(3/2)
(Normally you'd put a + C but we're doing a definite integral so it doesn't matter). At this point, you have two choices. You can either substitute back u = 2x + 1 and evaluate the 1/3 * (2x + 1)^(3/2) at x = 0 and x = 4. Or we could get the values of u that correspond to x = 0 and x = 4 and plug in that. I'll do the latter. When x = 0, we have u = 2(0) + 1 = 1, and when x = 4, we have u = 2(4) + 1 = 9, so we'll be evaluating 1/3 * u^(3/2) at u = 1 and u = 9:
1/3 * 9^(3/2) - 1/3 * 1^(3/2)
= 1/3 * 27 - 1/3
So 26/3 is the answer.
I hope this helps!