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 $\tuple {2, 0}$.
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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XV}$: $2$