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

∫110 ⁣x2−5x dx\$$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.∫16 ⁣x2−5x dx−∫106 ⁣x2−5x dx\$$int_1^6$$\!\sqrt{$$x^2-5x$$}\,dx-\int_{10}^6\!\sqrt{$$x^2-5x$$}\,dx

∫110 ⁣x2−5x dx\$$int_1$$^{10}\!\sqrt{$$x^2-5x$$}\,dx

∫101 ⁣x2−5x dx\int_{10}^1\!\sqrt{$$x^2-5x$$}\,dx

∫69 ⁣x2−5x dx\$$int_6^9$$\!\sqrt{$$x^2-5x$$}\,dx

∫96 ⁣x2−5x dx\$$int_9^6$$\!\sqrt{$$x^2-5x$$}\,dx

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0. Functions7h 55m

Introduction to Functions
18m

Piecewise Functions
10m

Properties of Functions
9m

Common Functions
1h 8m

Transformations
5m

Combining Functions
27m

Exponent rules
32m

Exponential Functions
28m

Logarithmic Functions
24m

Properties of Logarithms
36m

Exponential & Logarithmic Equations
35m

Introduction to Trigonometric Functions
38m

Graphs of Trigonometric Functions
44m

Trigonometric Identities
47m

Inverse Trigonometric Functions
48m

1. Limits and Continuity2h 2m

Introduction to Limits
41m

Finding Limits Algebraically
40m

Continuity
40m

2. Intro to Derivatives1h 33m

Tangent Lines and Derivatives
30m

Derivatives as Functions
14m

Basic Graphing of the Derivative
23m

Differentiability
26m

3. Techniques of Differentiation3h 18m

Basic Rules of Differentiation
39m

Higher Order Derivatives
12m

Product and Quotient Rules
55m

Derivatives of Trig Functions
40m

The Chain Rule
49m

4. Applications of Derivatives2h 38m

Motion Analysis
36m

Implicit Differentiation
27m

Related Rates
59m

Linearization
13m

Differentials
21m

5. Graphical Applications of Derivatives6h 2m

Intro to Extrema
23m

Finding Global Extrema
50m

The First Derivative Test
1h 5m

Concavity
1h 1m

The Second Derivative Test
31m

Curve Sketching
44m

Applied Optimization
1h 25m

6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m

Derivatives of Exponential & Logarithmic Functions
1h 52m

Logarithmic Differentiation
20m

Derivatives of Inverse Trigonometric Functions
24m

7. Antiderivatives & Indefinite Integrals1h 26m

Antiderivatives
17m

Indefinite Integrals
25m

Integrals of Trig Functions
15m

Initial Value Problems
27m

8. Definite Integrals4h 44m

Estimating Area with Finite Sums
42m

Riemann Sums
1h 21m

Introduction to Definite Integrals
22m

Fundamental Theorem of Calculus
37m

Average Value of a Function
21m

Substitution
1h 19m

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 31m

Integrals of Exponential Functions
40m

Integrals Involving Logarithmic Functions
58m

Integrals Involving Inverse Trigonometric Functions
52m

12. Techniques of Integration7h 41m

Integration by Parts
2h 32m

Partial Fractions
4h 29m

Improper Integrals
39m

13. Intro to Differential Equations2h 55m

Basics of Differential Equations
48m

Slope Fields
19m

Euler's Method
22m

Separable Differential Equations
1h 25m

14. Sequences & Series5h 36m

Sequences
1h 22m

Review of Factorials
11m

Series
44m

Convergence Tests
3h 17m

15. Power Series2h 19m

Introduction to Power Series
1h 9m

Taylor Series & Taylor Polynomials
1h 10m

16. Parametric Equations & Polar Coordinates7h 58m

Parametric Equations
1h 6m

Calculus with Parametric Curves
1h 13m

Polar Coordinates
2h 5m

Calculus in Polar Coordinates
55m

Conic Sections
2h 36m

8. Definite Integrals

Introduction to Definite Integrals

Calculus

8. Definite Integrals

Introduction to Definite Integrals

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

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

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

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

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

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

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.