Definite integrals: If int[-2,6]f(x)dx=-6, int[2,6]f(x)dx=16, int[-2,6]g(x)dx=4, find

[Jack.H]

If \(\displaystyle \displaystyle \int_{-2}^6\, f(x)\, dx\, =\, -6,\) \(\displaystyle \displaystyle \int_2^6\, f(x)\, dx\, =\, 16,\) and \(\displaystyle \displaystyle \int_{-2}^6\, g(x)\, dx\, =\, 4,\) find the following integrals:

. . . . .\(\displaystyle \displaystyle \mbox{1. }\, \int_{-2}^2\, f(x)\, dx\)

. . . . .\(\displaystyle \displaystyle \mbox{2. }\, \int_{-2}^6\, \left[g(x)\, +\, 2\right]\, dx\)

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I am not very good at integrals and don't really understand them, so I'm not very sure how I should approach this question. Could anyone explain to me how I should be approaching this question?

 

[tkhunny]

If \(\displaystyle \displaystyle \int_{-2}^6\, f(x)\, dx\, =\, -6,\) \(\displaystyle \displaystyle \int_2^6\, f(x)\, dx\, =\, 16,\) and \(\displaystyle \displaystyle \int_{-2}^6\, g(x)\, dx\, =\, 4,\) find the following integrals:

. . . . .\(\displaystyle \displaystyle \mbox{1. }\, \int_{-2}^2\, f(x)\, dx\)

. . . . .\(\displaystyle \displaystyle \mbox{2. }\, \int_{-2}^6\, \left[g(x)\, +\, 2\right]\, dx\)

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I am not very good at integrals and don't really understand them, so I'm not very sure how I should approach this question. Could anyone explain to me how I should be approaching this question?

Well, get better at it!

You need this: \(\displaystyle For\;a < b < c,\;\int\limits_{a}^{c}f(x)\;dx = \int\limits_{a}^{b}f(x)\;dx + \int\limits_{b}^{c}f(x)\;dx \)

This thing you should have been given in class or in your text.

 

[HallsofIvy]

In general, b does not have to be between a and c- in that case, at least one of the integrals will be negative.

 

[Steven G]

In general, b does not have to be between a and c- in that case, at least one of the integrals will be negative.

Exactly what I was going to say.

 

[Steven G]

View attachment 9387

I am not very good at integrals and don't really understand them, so I'm not very sure how I should approach this question. Could anyone explain to me how I should be approaching this question?

You also need to know that if you switch the limits of integration, then the result of the definite integral will have a sign change.

 

[tkhunny]

In general, ...

Exactly, ...

And we are not working generally. We are solving the f(x) problem.

Usually, I'm the one muddying the waters.