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Let us have a certain function like $f(x)=x$

In the world of integration, can I say that $\int_{0}^bxdx=\frac{1}{2}(b)(f(b))$? Because if you look at graph of function x, it would look like something 45° right triangle (though we can probably change its angle using the trig func $\tan$ and multiplying it by x like this $f(x)=x\tan(s)$ or in integration $\int_{0}^bx\tan(s)dx=\frac{1}{2}(b)(b\tan(s))$ or making it more complicated $\int_{a}^bx\tan(s°)dx=(\frac{1}{2}(b)(b\tan(s°)))-\frac{1}{2}(a)(f(a))$ where s is your/the preferred angle.

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    $\begingroup$ Yes, but this will only work if you’re integrating lines through the origin, of course. $\endgroup$ – symplectomorphic Jun 11 at 5:33
  • $\begingroup$ Further to @symplectomorphic's comment, an integral under a line, neither end on the $y$-axis, is a trapezium. This is the basis of the trapezium rule. $\endgroup$ – J.G. Jun 11 at 5:54

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