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int_{4}^{2}=-int_{1}^{4}-int_{0}^{1}+int_{0}^{2}
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well this is the original question .. Assuming that ∫[0,1,,] f(x)dx=3 ∫[0,2] f(x)dx=5 ∫[1,4] f(x)dx=8 What is ∫[4,2] f(x)dx?
and im not sure how to go about solving it
\[ \int_0^1f(x)dx=3, \int_0^2f(x)dx=5, \int_1^4f(x)dx=8, \int_4^2f(x)dx=? \] Is that the problem? If so: \[ \int_4^2f(x)dx=-\int_2^4f(x)dx=-\left( \int_1^4f(x)dx+\int_0^1f(x)dx-\int_0^2f(x)dx \right) \]
So yeah, you were correct, I just misunderstood what you were saying.
now how do i solve that part
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Just substitute in the numbers that it told you each of those integrals were equal to.
so i plug in 3 to the first integral? a
You plug in 3 for \(\int_0^1f(x)dx\), etc.
thanks
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