Evaluate the following integral: ∫ 0 1 ( 2 x 3 − x 2 + 4 x ) d x \int_0^1\!\left(2x^3-x^2+4x\right)\,dx ∫ 0 1 ( 2 x 3 − x 2 + 4 x ) d x

give this problem a try. So here we're asked to evaluate the integral from zero to one of this function right here. So we have this definite integral which we can go ahead and use the fundamental theorem of calculus part two to solve. Now what I'm first going to do is find the antiderivative of this piece of the function, and I can do that by finding the antiderivative of each individual portion according to the sum and difference rule that we learned for indefinite integrals, which these also apply to definite integrals as well. So go ahead and start right here for this integral of two x cubed. So that's going to give me using the power rule, have two x to the fourth power over four. Then we're going to have minus, and then using the power rule here, we'll have x cubed over three. Then we'll go ahead and add four x squared over two, taking the infinite derivative here. And then we're going to bound this whole thing from zero to one. So this is how we can deal with this integral. Now what I'm gonna do from here is reduce the fractions that I see inside. So I can divide this and reduce this fraction here that's going to give me x to the fourth power over two. Then I can go ahead and this fraction is already as simplified as it will get x cubed over three. Then we'll have plus and then I can go ahead and divide this four by two to give me two x squared. Then all of that's going to be bounded from zero to one. And what I need to do is apply my high bound and low bound respectively. So we're gonna go ahead and start with the high bound. So doing this, we're going to take this one here and plug it in for every place I see x. So we're gonna have one to the fourth power over two minus, and we'll have one cubed over three, plus then we're going to have two times one squared. This is what we get for this high bound. What I need to go ahead and do is plug in the low bound. So plugging in the low bound, well, I'm gonna do that over here. So we're going to have, this zero plugged in for every number. Now I could actually take zero and plug it in everywhere, but there's something I notice. Every place that I see x, that's just gonna multiply all out to be zero. So I don't even need to really worry about this low bound. I can just say the whole thing's gonna end up being zero just to save myself some time. So doing this, all we really need to do is evaluate this left side here. Well, we have one to the fourth power over two, which is one half. Then we have one cubed over three, which is one third. Then we have plus, and then two times one squared, which is two. Now at this point, I need to get a common fraction here, And the common number that I think is gonna get all these fractions the same is gonna be six. So if I go ahead and focus on this side, if I multiply the bottom and top by three, that's gonna give me six. We're gonna have three over six. It's the same thing as one half. And then if I multiply the top and bottom by two for this fraction, it's gonna give me two over six. And we're gonna have plus and then need to multiply this by six because that's gonna give me 12 over six. Right? Because, again, 12 divided by six gives us two, and then we reduce these fractions. We get these two as well. So we went ahead and got the common denominators here so we can just add the numerators across. So we have three minus two, which is one, and then plus 12, which is 13. So we get 13 over six as our final result. And that is the solution to this problem and how you can solve this definite integral. So hope you found this video helpful, and let's move on.

Evaluate the following integral: ∫ 0 1 ⁣ ( 2 x 3 − x 2 + 4 x ) d x \int_0^1\!\left(2x^3-x^2+4x\right)\,dx ∫ 0 1 ( 2 x 3 − x 2 + 4 x ) d x

− 11 6 -\frac{11}{6} − 6 11

8. Definite Integrals

Fundamental Theorem of Calculus

Business Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

Struggling with Business Calculus?

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Multiple Choice

Evaluate the following integral:
$\int_0^1\!\left(2x^3-x^2+4x\right)\,dx$

$\frac{13}{6}$

$\frac16$

$\frac{17}{6}$

$-\frac{11}{6}$

Verified step by step guidance

Step 1: Recognize that the problem involves evaluating a definite integral. The integral is given as ∫₀¹ (2x³ - x² + 4x) dx. This means we need to find the antiderivative of the function 2x³ - x² + 4x and then evaluate it at the bounds 0 and 1.

Step 2: Apply the power rule for integration to each term of the polynomial. The power rule states that ∫xⁿ dx = (xⁿ⁺¹) / (n+1) + C, where n ≠ -1. Using this rule, integrate each term: ∫2x³ dx = (2x⁴) / 4, ∫-x² dx = (-x³) / 3, and ∫4x dx = (4x²) / 2.

Step 3: Combine the results of the integration. The antiderivative of the function is: (2x⁴) / 4 - (x³) / 3 + (4x²) / 2. Simplify each term: (2x⁴) / 4 = (x⁴) / 2, (4x²) / 2 = 2x². So the antiderivative becomes: (x⁴) / 2 - (x³) / 3 + 2x².

Step 4: Evaluate the antiderivative at the upper and lower bounds of the integral. Substitute x = 1 into the antiderivative to find the upper bound value, and substitute x = 0 to find the lower bound value. The definite integral is the difference between these two values: [(1⁴) / 2 - (1³) / 3 + 2(1²)] - [(0⁴) / 2 - (0³) / 3 + 2(0²)].

Step 5: Simplify the expression obtained in Step 4. The terms involving x = 0 will all equal 0, so only the terms from x = 1 contribute to the result. Simplify the remaining terms to find the final value of the definite integral.