Question

1. given $\int_{0}^{3} f(x) d x=3, \int_{3}^{5} f(x) d x=-1, \int_{2}^{5} f(x) d x=2, \int_{2}^{5} g(x) d x=7$, find

(i) $\int_{2}^{5}(3 f(x)-2 g(x)) d x$,
(iii) $\int_{2}^{3} f(x) d x$,
(ii) $\int_{0}^{5} 3 f(x) d x$,
(iv) $\int_{2}^{0} f(x) d x$.

Explanation:

Step1: Use integral properties

For (i) $\int_{2}^{5}(3f(x)-2g(x))dx = 3\int_{2}^{5}f(x)dx-2\int_{2}^{5}g(x)dx$.
Given $\int_{2}^{5}f(x)dx = 2$ and $\int_{2}^{5}g(x)dx = 7$.
Substitute values: $3\times2-2\times7$.

Step2: Calculate the result

$3\times2-2\times7=6 - 14=-8$.

For (ii) $\int_{0}^{5}3f(x)dx = 3\int_{0}^{5}f(x)dx$.
Since $\int_{0}^{5}f(x)dx=\int_{0}^{3}f(x)dx+\int_{3}^{5}f(x)dx$, and $\int_{0}^{3}f(x)dx = 3$, $\int_{3}^{5}f(x)dx=-1$.
Then $\int_{0}^{5}f(x)dx=3+( - 1)=2$.
So $3\int_{0}^{5}f(x)dx=3\times2 = 6$.

For (iii) $\int_{2}^{3}f(x)dx=\int_{2}^{5}f(x)dx-\int_{3}^{5}f(x)dx$.
Substitute $\int_{2}^{5}f(x)dx = 2$ and $\int_{3}^{5}f(x)dx=-1$.
We get $2-( - 1)=3$.

For (iv) $\int_{2}^{0}f(x)dx=-\int_{0}^{2}f(x)dx$.
And $\int_{0}^{2}f(x)dx=\int_{0}^{5}f(x)dx-\int_{2}^{5}f(x)dx$.
Since $\int_{0}^{5}f(x)dx = 2$ and $\int_{2}^{5}f(x)dx = 2$.
Then $\int_{0}^{2}f(x)dx=2 - 2=0$.
So $\int_{2}^{0}f(x)dx=-0 = 0$.

Answer:

(i) $-8$; (ii) $6$; (iii) $3$; (iv) $0$