Evaluate \( \int_{0}^{2}|4 x-5| d x \).

1.	Evaluate \( \int_{0}^{2}\|4 x-5\| d x \).
Answer» \(\int\limits_0^2\|4x - 5\|dx\) \(\because\) \|4x - 5\| = \(\begin{cases}4x-5&; \quad x\geq\frac54\\-(4x-5) &;x\leq\frac54\end{cases}\) \(\therefore\) \(\int\limits_0^2\|4x-5\|dx=\int\limits_0^{5/4}-(4x-5)dx+\int\limits_{5/4}^2(4x-5)dx\) \(=[-\frac{4x^2}2+5x]_0^{5/4}+[\frac{4x^2}2-5x]_{5/4}^2\) \(=(-2\times(5/4)^2+5\times5/4)-0)+(2\times2^2-5\times2\) \(-2\times(5/4)^2+5\times5/4)\) \(=2\times\frac{25}4-4\times\frac{25}{16}+8-10\) \(=\frac{25}2-\frac{25}4-2\) \(\frac{25}4-2\) = \(\frac{25-8}4=\frac{17}4\).

Answer»

\(\int\limits_0^2|4x - 5|dx\)

\(\because\) |4x - 5| = \(\begin{cases}4x-5&; \quad x\geq\frac54\\-(4x-5) &;x\leq\frac54\end{cases}\)

\(\therefore\) \(\int\limits_0^2|4x-5|dx=\int\limits_0^{5/4}-(4x-5)dx+\int\limits_{5/4}^2(4x-5)dx\)

\(=[-\frac{4x^2}2+5x]_0^{5/4}+[\frac{4x^2}2-5x]_{5/4}^2\)

\(=(-2\times(5/4)^2+5\times5/4)-0)+(2\times2^2-5\times2\) \(-2\times(5/4)^2+5\times5/4)\)

\(=2\times\frac{25}4-4\times\frac{25}{16}+8-10\)

\(=\frac{25}2-\frac{25}4-2\)

\(\frac{25}4-2\) = \(\frac{25-8}4=\frac{17}4\).

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