$\int _{-2}^2\:f\left(x\right)dx+\int _2^5\:f\left(x\right)dx-\int _{-2}^{-1}\:f\left(x\right)dx$
I did it through basic arithmetic by adding and then subtracting the values of the top and the bottom with their corresponding top and bottom values, but that did not yield the correct answer.
Is there a special way to compute this that I am not aware of?
Any help?
$\endgroup$2 Answers
$\begingroup$The rules are: $$\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$$ $$\int_a^b f(x)dx = -\int_b^a f(x)dx.$$
$\endgroup$ $\begingroup$Original Equation: \begin{align*} \int_{-2}^{2}f(x)dx+\int_{2}^{5}f(x)dx-\int_{-2}^{-1}f(x)dx \end{align*} "Combine" first two integrals \begin{align*} \int_{-2}^{5}f(x)dx-\int_{-2}^{-1}f(x)dx\ \end{align*} "Flip" last integral and change from negative to positive \begin{align*} \int_{-2}^{5}f(x)dx+\int_{-1}^{-2}f(x)dx\ \end{align*} "Combine" Integrals \begin{align*} \int_{-1}^{5}f(x)dx. \end{align*}
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