Integration

Question

$\int_{-4}^{4}\left | x-3 \right |dx$

Answer 1

Here is the graph of this function : $\int_{-4}^{4}|x-3|$

Observing here we find that at $x=3$ function is $0$ and before that it is negative therefore we can work out integration of this function by removing absolute function as follows :

$\Rightarrow $ $\int_{-4}^{3}-(x-3)+\int_{3}^{4}(x-3)$

$\Rightarrow $ $\int_{-4}^{3}(3-x)\int_{3}^{4}(x-3)$

$\Rightarrow $ $[3x-\frac{x^2}{2}]_{-4}^{3}+[\frac{x^2}{2}-3x]_{3}^4$

$\Rightarrow \ 25$

Answer 2

$\int_{-4}^{3}(-x+3)+\int_{3}^{4}(x-3)$

Comments

why -4 to 3 r u dividing the integration?

not getting this point

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 -4 to 3 ...in this interval....x-3 is negative therefor mod(x-3) will open with negative sign...isn't it?

for remaining interval (x-3) is +ve ...therefor we will remove mod simply....