Is this inequality true and provable?

[3029298]

Homework Statement

My question is whether the following inequality can be proven.

Homework Equations

$$\left|\int_a^bg\left(x\right)dx-\int_a^bh\left(x\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$

The Attempt at a Solution

I tried to write down the inequality in the form of it's primitives, where $$G\left(x\right)$$ is the primitive of $$g\left(x\right)$$ and $$H\left(x\right)$$ is the primitive of $$h\left(x\right)$$. The inequality then becomes:

$$\left|G\left(b\right)-G\left(a\right)-H\left(b\right)+H\left(a\right)\right|\leq\left|G\left(b\right)-H\left(b\right)\right|-\left|G\left(a\right)-H\left(a\right)\right|$$

But what next, or are there other means of getting a proof?

 

[snipez90]

Assuming $$a \leq b$$ and f is continuous on the interval [a,b], then

$$\left|\int_a^bf\left(x\right)dx \right| \leq \int_a^b\left|f(x)\right|dx$$

which follows from the fact that $$f(x) \leq \left|f(x)\right|$$ and $$-f(x) \leq \left|f(x)\right|$$ and that

If f,g are both continuous on the interval [a,b] and $$f(x) \leq g(x)$$ for all x in the interval. Then

$$\int_a^b f(x)dx \leq \int_a^b g(x)dx$$

Rearranging and using the first inequality should give you the desired inequality.

 

[3029298]

Oh, I see it now, it is indeed not that difficult.

$$\left|\int_a^bg\left(x\right)dx-\int_a^bh\left(x\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$

If we rearrange:

$$\left|\int_a^b\left(g\left(x\right)-h\left(x\right)\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$

Substituting $$f\left(x\right)=g\left(x\right)-h\left(x\right)$$ and using the first formula of snipez90, we get the proof.

Thanks!