Evaluate the definite integral from 0 to 1 of
(x^2)sqrt(5x+6)dx
∫[0,1] x^2 √(5x+6) dx
That's almost x^5/2, but for that pesky 5x+6. So, let u=5x+6 and we have x = (u-6)/5:
∫[6,11] (u-6)^2/25 √u du/5
= 1/125 ∫[6,11] u^(5/2) - 12u^(3/2) + 36u^(1/2) du
= 1/125 (2/7 u^7/2 - 24/5 u^5/2 + 24u^3/2)[6,11]
= 2/4375 (576√6 - 1111√11)
= 1.03948
To evaluate the definite integral ∫[0,1] (x^2)sqrt(5x+6) dx, we can use the technique of integration by substitution. Here's how we can solve it step by step:
1. Start by making a substitution: Let u = 5x + 6. Differentiating both sides with respect to x gives du/dx = 5.
2. Rearrange the equation to solve for dx: dx = du/5.
3. Substitute the values of u and dx in terms of du into the integral. The new integral becomes:
∫[(0,1)] (x^2)sqrt(u) * (1/5) du.
4. Adjust the limits of integration based on the new variable. When x = 0, u = 6, and when x = 1, u = 11.
5. Rewrite the integral using the new variable: (1/5) * ∫[6,11] (x^2)sqrt(u) du.
6. Calculate the integral of u^(1/2): ∫[6,11] sqrt(u) du = (2/3) * u^(3/2) evaluated from 6 to 11.
7. Apply the limits of integration to get the final result: (1/5) * [(2/3) * (11^(3/2) - 6^(3/2))].
8. Compute the numerical value to get the final answer.
Keep in mind that these steps may vary depending on the form of the integral and the technique used.