Question

suppose that $f(x)$ is an even function which satisfies the following conditions:
(a) $int_{0}^{11} f(x) mathrm{d} x=15$
(b) $int_{-3}^{3} f(x) mathrm{d} x=18$
evaluate the following integral:
$$int_{3}^{11}(f(x)+10) mathrm{d} x.$$
$$int_{3}^{11}(f(x)+10) mathrm{d} x=$$ number (3 significant figures)

Explanation:

Step1: Use integral property

$$\int_{3}^{11}(f(x)+10)dx=\int_{3}^{11}f(x)dx+\int_{3}^{11}10dx$$

Step2: Use even - function property

Since \(f(x)\) is even, \(\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\). Given \(\int_{-3}^{3}f(x)dx = 18\), then \(\int_{0}^{3}f(x)dx=\frac{1}{2}\int_{-3}^{3}f(x)dx = 9\). Also, \(\int_{0}^{11}f(x)dx = 15\). So \(\int_{3}^{11}f(x)dx=\int_{0}^{11}f(x)dx-\int_{0}^{3}f(x)dx=15 - 9=6\)

Step3: Calculate \(\int_{3}^{11}10dx\)

Using \(\int_{a}^{b}kdx=k(b - a)\) (\(k = 10\), \(a = 3\), \(b = 11\)), we have \(\int_{3}^{11}10dx=10\times(11 - 3)=80\)

Step4: Sum the two integrals

\(\int_{3}^{11}(f(x)+10)dx=\int_{3}^{11}f(x)dx+\int_{3}^{11}10dx=6 + 80=86\)

Answer:

\(86\)