# Even and odd integrals

Find the definite integral

$$\int_{-2}^{2} \Big(2f(x) + 3g(x)\Big)dx$$

where $f(x)$ is an even function such that

$$\int_{0}^{2} f(x)dx = 3$$

and $g(x)$ is such that

$$\int_{-2}^{4} g(x)dx = -3 \ \ \text{and}\ \ \int_{2}^{4} g(x)dx = -6$$

• Hint: Split the integral up (addivitivity). Then split each of those integrals up again (property of integrals involving limits of integration). – user137731 Dec 14 '14 at 14:28

## 1 Answer

By the properties of the integral we have \begin{align*} \int_{-2}^{2} \Big(2f(x) + 3g(x)\Big)dx&=2\int_{-2}^2f(x)\,dx+3\int_{-2}^2g(x)\,dx\\ &=2\left[2\int_0^2f(x)\,dx\right]+3\int_{-2}^2g(x)\,dx\quad\text{Since }f\text{ is an even function}\\ &=2[2(3)]+3\left[\int_{-2}^4g(x)\,dx-\int_2^4g(x)\,dx\right]\\ &=2(6)+3[-3-(-6)]\\ &=12+3(3)\\ &=21 \end{align*}