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• Notes

$\int_{0}^{9} f(x) \mathrm{d} x=37 \quad, \quad \int_{0}^{9} g(x) \mathrm{d} x=16 \quad \text { Find } \int_{0}^{9}(2 f(x)+3 g(x)) \mathrm{d} x$

$\int_{0}^{9}(2 f(x)+3 g(x)) d x=\int_{0}^{9} 2 f(x) d x+\int_{0}^{4} 3 g(x) d x$

$=2 \int_{0}^{9} f(x) d x+3 \int_{0}^{9} g(x) d x$

$=2(37)+3(16)=122$

Use the properties of the integral to verify the Inequality: $-\frac{1}{3} \leq \int_{0}^{1} x^{2} \cos x \mathrm{d} x \leq \frac{1}{3}$

$-1 \leq \cos x \leq 1 \quad \quad x^{2} \geq 0$

$-x^{2} \leq x^{2} \cos x \leq x^{2}$

$-\int_{0}^{1} x^{2} d x \leq \int_{0}^{1} x^{2} \cos x d x \leq \int_{0}^{1} x^{2} d x$

$-\left[\frac{x^{3}}{3}\right]_{0}^{1} \leq \int_{0}^{1} x^{2} \cos x d x \leq\left[\frac{x^{3}}{3}\right]_{0}^{1}$

$-\left(\frac{1}{3}-0\right) \leq \int_{0}^{1} x^{2} \cos x d x \leq \frac{1}{3}-0$

$-1 / 3 \leq \int_{0}^{1} x^{2} \cos x d x \leq \frac{1}{3}$