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my work is as follows:

$$\int^6_0 f(x)dx = \int^2_0 f(x)dx - \int^6_0 f(x)dx$$

I then got, $7 - 15 = -8$

Therefore, $\int^6_0 f(x)dx = -8$

I think this is right but I'm not sure, so if someone can confirm or deny, that would be great, thanks!

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    $\begingroup$ Please use MathJax to format. $\endgroup$ – Saad May 8 '20 at 14:28
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    $\begingroup$ The equality you start with is not true. What did you mean to write? $\endgroup$ – lulu May 8 '20 at 14:31
  • $\begingroup$ given a<b<c then we have the intuitive formula $\int_a^c f(x) = \int_a^b f(x) + \int_b^c f(x)$ $\endgroup$ – Aladin May 8 '20 at 14:35
  • $\begingroup$ As a suggestion, it is easy to come up with step functions that meet the assumptions. Say $f(x)=\frac 72$ for $0≤x≤2$ and $f(x)=\frac {15}4$ for $2<x≤6$. You can then work out the answer for this example. That won't prove that the answer is always correct, but you could at least see that $-8$ is not correct. And knowing the correct answer is a great start. $\endgroup$ – lulu May 8 '20 at 14:36
  • $\begingroup$ Sometimes it can help to think about this in words: "The area under $f$ from $0$ to $2$ is 7, and the area under $f$ from $2$ to $6$ is 15. What's the area under $f$ from $0$ to $6$?" $\endgroup$ – user113102 May 8 '20 at 14:42
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$\int_0^6f(x)dx=\int_0^2f(x)dx+\int_2^6f(x)dx$

$\int_0^6f(x)dx=7+15=22$

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You can also look at what a definite integral means geometrically.

$\int_a^b f(x)dx$ is the area under the curve from $a$ to $b$. For example, for this curve, it would be the blue area.

enter image description here

Now, imagine there is a point $c$ between $a$ and $b$ and you know the area from $a$ to $c$ and the area from $c$ to $b. For example:

enter image description here

With these two areas, you know that the area from $a$ to $b$ must be $22 = 7 + 15$. Or, to put it in integral terms:

$\int_a^b f(x) = \int_a^c f(x) + \int_c^b f(x)$

$\int_a^c f(x) = 7$

$\int_c^b f(x) = 15$

Therefore

$\int_a^b f(x) = 7 + 15 = 22$

In the context of your original question: $a = 0, c = 2, b = 6$

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Well, it is not hard to notice that when $0<\text{a}<\text{b}$:

$$\int_0^\text{b}\text{y}\left(x\right)\space\text{d}x=\int_0^\text{a}\text{y}\left(x\right)\space\text{d}x+\int_\text{a}^\text{b}\text{y}\left(x\right)\space\text{d}x\tag1$$

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