# Suppose that p of x is continuous at all real numbers and satisfies the following equations....

## Question:

Suppose that {eq}p(x) {/eq} is continuous at all real numbers and satisfies the following equations.

{eq}\bullet {\displaystyle \int_2^5} p(x) dx =4 {/eq}

{eq}\bullet {\displaystyle \int_2^{10}} p(x) dx =13 {/eq}

{eq}\bullet {\displaystyle \int_{10}^{25}} p(x) dx =61 {/eq}

{eq}\bullet {\displaystyle \int_{20}^{25}} p(x) dx =36 {/eq}

What is the value of {eq}{\displaystyle \int_5^{20}} (3p(x)-4) dx {/eq}?

## Integration:

Integration is defined as area under the curve. Properties of integration:

{eq}1. \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx \\ 2. \int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx \ \ \ \ a< c < b\\ {/eq}

## Answer and Explanation:

{eq}\int_{5}^{10} p(x) = \int_{2}^{10} p(x) - \int_{2}^{5} p(x)\\ \int_{5}^{10} p(x) = 13 -4 = 9\\ \int_{10}^{20} p(x)= \int_{10}^{25} p(x) - \int_{20}^{25} p(x)\\ \int_{10}^{20} p(x)= 61 - 36 = 25\\ {\displaystyle \int_5^{20}} p(x) = \int_{5}^{10} p(x)+ \int_{10}^{20} p(x)\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 {\displaystyle \int_5^{20}} p(x) -4\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 (\int_{5}^{10} p(x)+ \int_{10}^{20} ) -4\\ {\displaystyle \int_5^{20}} p(x) (3p(x)-4) dx = 3 cdot (9 + 25) -4 = 102-4 = 98 {/eq} 