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

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.

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

8. Definite Integrals

Fundamental Theorem of Calculus

Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

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Multiple Choice

Evaluate the following integral:
$\int_{- 1}^{2} (x^{2} - 3 x + 2) d x$

-6

$27 over 6$

$25 over 6$

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Verified step by step guidance

Step 1: Identify the integral to be evaluated, which is ∫ from -1 to 2 of (x^2 - 3x + 2) dx.

Step 2: Find the antiderivative of the integrand. The antiderivative of x^2 is (1/3)x^3, the antiderivative of -3x is (-3/2)x^2, and the antiderivative of 2 is 2x.

Step 3: Combine these antiderivatives to form the antiderivative of the entire integrand: F(x) = (1/3)x^3 - (3/2)x^2 + 2x.

Step 4: Evaluate the definite integral by applying the Fundamental Theorem of Calculus, which involves calculating F(2) - F(-1).

Step 5: Substitute x = 2 and x = -1 into the antiderivative F(x) and subtract the results to find the value of the definite integral.

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Evaluate the following integral:∫−12(x2−3x+2)dx\int_{-1}^2\!\left(x^2-3x+2\right)\,dx

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Fundamental Theorem of Calculus practice set

Evaluate the following integral:
$\int_{- 1}^{2} (x^{2} - 3 x + 2) d x$