Write the two definite integrals subtracted below as a single integral.∫16x2−5xdx−∫106x2−5xdx\int_1^6\!\sqrt{x^2-5x}\,dx-\int_{10}^6\!\sqrt{x^2-5x}\,dx

∫ 1 10 ⁣ x 2 − 5 x d x \int_1^{10}\!\sqrt{x^2-5x}\,dx

Write the two definite integrals subtracted below as a single integral. ∫ 1 6 x 2 − 5 x d x − ∫ 10 6 x 2 − 5 x d x \int_1^6\!\sqrt{x^2-5x}\,dx-\int_{10}^6\!\sqrt{x^2-5x}\,dx

this problem, we're asked to write the two definite integrals subtracted below as a single integral. Now to start for this problem, I'm looking at the two integrals that I have here, and I notice something kind of interesting. We have the same function inside of both of these integrals, but notice that the bounds are wildly different. This bound goes from one to six, and this we can think of as a negative integral going from 10 to six. Now because I see this going from 10 to six, I'm actually going to rewrite this by switching the bounds and changing the sign. So what's gonna give me is we'll first have the integral from one to six, and it's gonna be the square root of x squared minus five x with respect to x. Then what I'm going to do is flip the bounds here. So we're going to get the negative, and then it's going to be minus, and then we're going to which we're gonna have minus the negative, and then it's gonna be the integral from, six to 10 flipping these bounds. We're gonna have the square root of x squared minus five x, all with respect to x. And since we're going ahead and making this whole thing negative, those are just gonna cancel. Meaning this is all going to come out to the integral from one to six. We'll just keep this first portion square root of x squared minus five x d x plus the integral from six to 10 of the square root of x squared minus five x, all with respect to x. Now notice something that happens here when I set up the integrals like this. Notice that this six here matches with that six right there. And one of the things I notice is six is a number that's between one and ten. And there's a rule that we discuss when this happens. Whenever this happens, you can write this as one big integral that goes from the lowest bound one to the highest bound of 10. And then since this function is the same right here, we're just gonna keep the square root of x squared minus five x inside, and then that's all with respect to x. And notice how we've managed to take this integral that was originally two separate definite integrals and combined it into one. And that right there would be the simplified solution to our problem, and that is how we can combine this into a single integral. So that's how you can solve these types of problems where you have to apply the rules of integration. Hope you found this video helpful, and let's move on and get some more practice.

Write the two definite integrals subtracted below as a single integral. ∫ 1 6 ⁣ x 2 − 5 x d x − ∫ 10 6 ⁣ x 2 − 5 x d x \int_1^6\!\sqrt{x^2-5x}\,dx-\int_{10}^6\!\sqrt{x^2-5x}\,dx

∫ 1 10 ⁣ x 2 − 5 x d x \int_1^{10}\!\sqrt{x^2-5x}\,dx

∫ 10 1 ⁣ x 2 − 5 x d x \int_{10}^1\!\sqrt{x^2-5x}\,dx

∫ 6 9 ⁣ x 2 − 5 x d x \int_6^9\!\sqrt{x^2-5x}\,dx

∫ 9 6 ⁣ x 2 − 5 x d x \int_9^6\!\sqrt{x^2-5x}\,dx

8. Definite Integrals

Introduction to Definite Integrals

Calculus

8. Definite Integrals

Introduction to Definite Integrals

Struggling with Calculus?

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Multiple Choice

Write the two definite integrals subtracted below as a single integral.
$\int_{1}^{6} \sqrt{x^{2} - 5 x} d x - \int_{10}^{6} \sqrt{x^{2} - 5 x} d x$

$\int_{10}^{1} \sqrt{x^{2} - 5 x} d x$

$\int_{1}^{10} \sqrt{x^{2} - 5 x} d x$

$\int_{6}^{9} \sqrt{x^{2} - 5 x} d x$

$\int_{9}^{6} \sqrt{x^{2} - 5 x} d x$

Verified step by step guidance

First, understand that the problem involves combining two definite integrals into a single integral. The integrals are: ∫ from 1 to 6 of √(x² - 5x) dx and ∫ from 10 to 6 of √(x² - 5x) dx.

Recognize that the second integral, ∫ from 10 to 6, is actually the negative of ∫ from 6 to 10. This is because reversing the limits of integration changes the sign of the integral.

Rewrite the second integral with reversed limits: ∫ from 6 to 10 of √(x² - 5x) dx. This changes the sign, so the original subtraction becomes addition.

Combine the two integrals: ∫ from 1 to 6 of √(x² - 5x) dx + ∫ from 6 to 10 of √(x² - 5x) dx.

Finally, express the combined integral as a single integral: ∫ from 1 to 10 of √(x² - 5x) dx. This represents the total area under the curve from x = 1 to x = 10.

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