1. ## Integration

Let $\displaystyle G(x)=\int_{0}^{x} f(x)dx$

The graph of f(x) is under the x axis from [-2,0]

Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because

$\displaystyle G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx$ = -(-area)=positive area

But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks

2. Originally Posted by Linnus

Let $\displaystyle G(x)=\int_{0}^{x} f(x)dx$

The graph of f(x) is under the x axis from [-2,0]

Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because

$\displaystyle G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx$ = -(-area)=positive area Mr F says: This is G(-2), NOT G(x).

But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks
Correct, it is indeed decreasing over the interval $\displaystyle -2 \leq x \leq 0$.

Consider:

$\displaystyle G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0$.

$\displaystyle G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0$.

But clearly G(-2) > G(-1) .......

3. Originally Posted by mr fantastic
Correct, it is indeed decreasing over the interval $\displaystyle -2 \leq x \leq 0$.

Consider:

$\displaystyle G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0$.

$\displaystyle G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0$.

But clearly G(-2) > G(-1) .......
Ah...that makes sense.