Step 1

Given:

\(\int_{-2}^{2}f(x)dx=4\)

\(\int_{2}^{5}f(x)dx=3\)

\(\int_{-2}^{5}g(x)dx=2\)

Step 2

Using properties of Definite Integrals:

\(\int_{a}^{b}[f(x)\pm g(x)]dx=\int_{a}^{b}f(x)dx\pm \int_{a}^{b}g(x)dx\)

\(\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\int_{a}^{b}f(x)dx\ \ \ [c\in(a,b)]\)

\(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\)

Step 3

Now,

\(\int_{-2}^{5}f(x)dx=\int_{-2}^{2}f(x)dx+\int_{2}^{5}f(x)dx=4+3=7\)

\(\int_{-2}^{5}g(x)dx=2\)

\(\therefore\) On interval \(-2\leq x\leq 5\)

\(f(x)\geq g(x)\)

False Statement

Given:

\(\int_{-2}^{2}f(x)dx=4\)

\(\int_{2}^{5}f(x)dx=3\)

\(\int_{-2}^{5}g(x)dx=2\)

Step 2

Using properties of Definite Integrals:

\(\int_{a}^{b}[f(x)\pm g(x)]dx=\int_{a}^{b}f(x)dx\pm \int_{a}^{b}g(x)dx\)

\(\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\int_{a}^{b}f(x)dx\ \ \ [c\in(a,b)]\)

\(\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx\)

Step 3

Now,

\(\int_{-2}^{5}f(x)dx=\int_{-2}^{2}f(x)dx+\int_{2}^{5}f(x)dx=4+3=7\)

\(\int_{-2}^{5}g(x)dx=2\)

\(\therefore\) On interval \(-2\leq x\leq 5\)

\(f(x)\geq g(x)\)

False Statement