# How do you find the definite integral of #x^2 * sqrt (x^3 + 1) * dx# in the interval #[1,2]#?

Hopefully this helps!

By signing up, you agree to our Terms of Service and Privacy Policy

To find the definite integral of (x^2 \cdot \sqrt{x^3 + 1}) with respect to (x) over the interval ([1,2]), you can use integration by substitution. Let (u = x^3 + 1), then (du = 3x^2 , dx). Rearranging, we have (dx = \frac{du}{3x^2}).

Substituting these into the integral, we get:

[\int_{1}^{2} x^2 \sqrt{x^3 + 1} , dx = \frac{1}{3} \int_{2}^{9} \sqrt{u} , du.]

Integrating (\sqrt{u}) with respect to (u), we get:

[\frac{1}{3} \left[\frac{2}{3}u^{\frac{3}{2}}\right]_{2}^{9} = \frac{2}{9}(9^{\frac{3}{2}} - 2^{\frac{3}{2}}).]

Finally, evaluating this expression, we find:

[\frac{2}{9}(27 - 2\sqrt{2}).]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7