Area under Curve/Examples/(x - 1) (x - 2)

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Example of Use of Area under Curve

The area between the $x$-axis and the curve $y = \paren {x - 1} \paren {x - 2}$ is $\dfrac 1 6$.


Proof

Let $\AA$ be the area in question.

The curve $y = \paren {x - 1} \paren {x - 2}$ intercepts the $x$-axis where $y = 0$.

That is, where $x - 1 = 0$ and $x - 2 = 0$, which is $\tuple {1, 0}$ and $\tuple {2, 0}$.


Area-under-Curve-(x-1)(x-2).png


Thus from Area under Curve we need to evaluate the definite integral:

$\AA = \ds \int_1^2 \paren {x - 1} \paren {x - 2} \rd x$

Between those limits, $y$ is negative.

Hence we expect a negative result, which must then be made positive.


So:

\(\ds \AA\) \(=\) \(\ds \size {\int_1^2 \paren {x - 1} \paren {x - 2} \rd x}\)
\(\ds \) \(=\) \(\ds \size {\int_1^2 \paren {x^2 - 3 x + 2} \rd x}\)
\(\ds \) \(=\) \(\ds \size {\intlimits {\dfrac {x^3} 3 - \dfrac {3 x^2} 2 + 2 x} 1 2}\) Primitive of Power
\(\ds \) \(=\) \(\ds \size {\paren {\dfrac {2^3} 3 - \dfrac {3 \times 2^2} 2 + 2 \times 2} - \paren {\dfrac {1^3} 3 - \dfrac {3 \times 1^2} 2 + 2 \times 1} }\) Definition of Definite Integral
\(\ds \) \(=\) \(\ds \size {\paren {\dfrac 8 3 - 6 + 4} - \paren {\dfrac 1 3 - \dfrac 3 2 + 2} }\) evaluation
\(\ds \) \(=\) \(\ds \size {\dfrac 7 3 + \dfrac 3 2 - 4}\) evaluation
\(\ds \) \(=\) \(\ds \size {\dfrac {14 + 9 - 24} 6}\) evaluation
\(\ds \) \(=\) \(\ds \size {-\dfrac 1 6}\) evaluation
\(\ds \) \(=\) \(\ds \dfrac 1 6\) evaluation

$\blacksquare$


Sources