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Mathematics 20 Online
OpenStudy (anonymous):

Evaluate the integral ∫(x2 + 4x -2)dx from x = 1 to x = 4.

OpenStudy (anonymous):

A. 16 B. 25 C.45 D. 49

OpenStudy (luigi0210):

Integrate it first

OpenStudy (anonymous):

what does that mean?

OpenStudy (luigi0210):

You don't know what integrate means? ._.

OpenStudy (anonymous):

......no. im not good at this stuff

OpenStudy (mathstudent55):

\( \int x^n dn = \dfrac{x^{n + 1}}{n+1} + C \)

OpenStudy (luigi0210):

It means work backwards to the original equation. But looks like mathstudent is cooking something up, so I'll wait~

OpenStudy (mathstudent55):

Integration is the opposite of differentiation. If you start with a function and you integrate it you get a new function. If you differentiate the new function, you get your original function back.

OpenStudy (anonymous):

so i just plug everything in?

OpenStudy (anonymous):

and if so, what is n?

OpenStudy (mathstudent55):

For example: \( \int (x^2 + 3x + 1)dx = \dfrac{x^3}{3} + \dfrac{3x^2}{2} + x + C\) Now differentiate the new function: \( \dfrac{d}{dx} (\dfrac{x^3}{3} + \dfrac{3x^2}{2} + x + C) \) = \(= \dfrac{3x^2}{3} + \dfrac{6x}{2} + 1\) \(= x^2 + 3x + 1\) As you can see, when you integrate and differentiate, you get back to the same starting point.

OpenStudy (mathstudent55):

Follow the integration part of my example.

OpenStudy (anonymous):

how did u get 3 and 2 as denominators?

OpenStudy (mathstudent55):

Look at the formula for integrating x^n: \(\int x^n dn = \dfrac{x^{n + 1}}{n+1} + C\)

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