Definite Integral Calculation

Hi Recep, let's solve this definite integral together. It looks a bit complex at first, but we will break it down step by step.

The problem asks us to evaluate the integral from zero to one of the polynomial three x cubed minus x squared plus two x minus four, all divided by the square root of x squared minus three, plus two.

Before we start integrating, we must check the domain of the integrand. Look at the term inside the square root: x squared minus three.

The expression inside a square root must be greater than or equal to zero for the result to be a real number. So, x squared minus three must be at least zero.

Adding three to both sides, we see that x squared must be greater than or equal to three.

This implies that x must be greater than or equal to the square root of three, or less than or equal to negative square root of three.

Now, let's look at the limits of our integral. We are integrating from x equals zero to x equals one.

Wait a second. Our interval for integration is entirely between zero and one, but the square root in the denominator is only defined for x values outside of the range between negative root three and positive root three.

Note that the square root of three is approximately one point seven three two. Since our interval is inside this range, the term x squared minus three will be negative.