- #1

- 244

- 0

Is that [tex] f(b) - f(a) = f'(c)(b-a) [/tex] or am I thinking of something else?

- Thread starter tandoorichicken
- Start date

- #1

- 244

- 0

Is that [tex] f(b) - f(a) = f'(c)(b-a) [/tex] or am I thinking of something else?

- #2

- 976

- 0

[tex]\frac{\int_a^b f(x)\,dx}{b-a}}[/tex]

Just remember that you're turning the area under the curve into a rectangle of base (b - a). The height is the average value.

cookiemonster

- #3

- 244

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okay, the actual problem now becomes

[tex] \frac{\int_{-1}^{1} e^{-x^2} \,dx}{-2} [/tex]

any takers?

[tex] \frac{\int_{-1}^{1} e^{-x^2} \,dx}{-2} [/tex]

any takers?

- #4

- 976

- 0

The expression can be simplified at most to:

[tex]\int_0^1 e^{-x^2}\,dx[/tex]

Which must be integrated numerically. Mathematica gives:

[tex]\int_0^1 e^{-x^2}\,dx = .746824[/tex]

cookiemonster

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