How would I find the total area? On which steps?

Question

How to find the total area between $f(x) = x^2 - x$ and the $x-$axis over $[-1,2]$?

Answer 1

You have the upper bound $x=2$ and lower bound $x=-1$
simply integrate $$\int_{-1}^2y.dx$$ $$\int_{-1}^2x^2-x.dx$$ $$\int_{-1}^1x-x^2.dx+\int_1^2x^2-x.dx$$ should give you the answer

Answer 2

$$ y(x) = x^{2} - x $$

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Simplest interpretation:

$$ \int \left( x^{2} - x \right) dx = \frac{x^3}{3}-\frac{x^2}{2} $$ Therefore $$ \int_{-1}^{2} \left( x^{2} - x \right) dx = % \frac{1}{3}x^3-\frac{1}{2}x^2 \Bigg\lvert_{-1}^{2} = \left( \frac{8}{3} - \frac{4}{2} \right) - \left( \frac{-1}{3} - \frac{1}{2} \right) = \frac{3}{2} $$

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The Wainfleet Conjecture:

@DanielWainfleet makes an interesting point. Does the OP want this integral?

Solve for the roots: $$ x^{2}-x = x(1-x) = 0 \qquad \Rightarrow \qquad x=0, \ x = 1 $$ Parse the integral $$ \int_{-1}^0 f(x) \, dx - \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6} $$

Answer 3

The OP wants the area bounded by $y(x) = x^2 - x$ and the $x$-axis. Therefore, does the OP mean

$$\int_{-1}^0 (x^2-x) dx+\int_1^2 (x^2-x) dx=\frac{5}{3}?$$