Evaluate the following integral: ∫ − 1 2 ( x 2 − 3 x + 2 ) d x \int_{-1}^2\!\left(x^2-3x+2\right)\,dx ∫ − 1 2 ( x 2 − 3 x + 2 ) d x

this problem, we are asked to evaluate the following indefinite integral. Now I notice here that we're dealing with this function and we're integrating this from negative one to two, so we're dealing with this definite integral with these bounds, and we can use the fundamental theorem of calculus to solve this. Now the first thing I'm going to do is find the antiderivative of this function. And using the sum and difference rules, I can integrate each one of these pieces separately. Now to avoid writing this all out as three separate integrals, we're just gonna go ahead and go piece by piece here. So I'll start with the integral of x squared or the antiderivative, which is gonna come out to x cubed over three using the power rule. Then we'll have minus negative use the power rule here, which will give me three x squared over two. Then we'll have the antiderivative of two, which will give me two x. And all of this is going to be bounded from from negative one to two since these are the bounds we have on our integral. So we've gone ahead and taken the antiderivative of this function. And from here, what I need to do is plug in my bounds. So I'm gonna start with my high bound of two. So plugging that into my function here, we're going to just put this in for every place we see x. So I'm going to have two cubed over three minus, and that's gonna be three times two squared over two plus, and then we're going to have two times two. All of that is gonna be this first bound. Then we're going to subtract off everything with the lower bound plugged in. It's going to be negative one cubed over three. Then we're gonna have minus three times negative one squared over two, and it's gonna be plus two times negative one. And then this is gonna be everything with our low bound plugged in. Now obviously, we got a lot of numbers that we're dealing with here. So let's see how we can simplify things. Now first off, I'm going to simplify everything here on the left side here. If you go ahead and deal with all these numbers, use subtract, multiply, simplify these fractions. You should get two thirds as the results for everything right here. And then if you go ahead and do the same process for everything we have on the right side for all of these numbers, you should get negative 23 over six as the result on the right side there for these values that we have in the brackets. So this is what we end up getting. And what we can do is cancel this negative sign to get a positive. So that's going to give me two thirds plus 23 over six, which I can also write as four over six plus 23 over six. This allows me to get these like denominators. 23 plus four comes up to 27, so we get 27 over six as the final result and the solution to this problem. So this is how you can solve these problems with we're dealing with definite integrals. When you have these polynomials, you first just take the anti derivative of this function, and then you apply the bounds. So hope you found this video helpful, and let's move on.

8. Definite Integrals

Fundamental Theorem of Calculus

Business Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Evaluate the following integral:
$\int_{-1}^2\!\left(x^2-3x+2\right)\,dx$

-6

$\frac{27}{6}$

$\frac{25}{6}$

Verified step by step guidance

Step 1: Recognize that the problem involves evaluating a definite integral. The integral is given as ∫_{-1}^2 (x^2 - 3x + 2) dx. This means we need to find the antiderivative of the function x^2 - 3x + 2 and then evaluate it at the bounds -1 and 2.

Step 2: Find the antiderivative of the function x^2 - 3x + 2. To do this, integrate each term separately: ∫x^2 dx, ∫(-3x) dx, and ∫2 dx. Use the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C for n ≠ -1.

Step 3: Apply the power rule to each term: ∫x^2 dx = (x^3)/3, ∫(-3x) dx = (-3x^2)/2, and ∫2 dx = 2x. Combine these results to get the antiderivative: F(x) = (x^3)/3 - (3x^2)/2 + 2x.

Step 4: Use the Fundamental Theorem of Calculus to evaluate the definite integral. Substitute the upper limit (x = 2) and the lower limit (x = -1) into the antiderivative F(x). Compute F(2) and F(-1) separately.

Step 5: Subtract the value of the antiderivative at the lower limit from the value at the upper limit: ∫_{-1}^2 (x^2 - 3x + 2) dx = F(2) - F(-1). Simplify the result to find the value of the definite integral.