Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\int_{1}^{4}|\mathrm{x}-3| \mathrm{d} \mathrm{x}$ is equal to
Options:
Solution:
2171 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{2}$
$\begin{aligned}
& \int_{1}^{4}|\mathrm{x}-3| \mathrm{dx}=\int_{1}^{3}-(\mathrm{x}-3) \mathrm{dx}+\int_{3}^{4}(\mathrm{x}-3) \mathrm{dx} \\
&=-\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{1}^{3}+\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{3}^{4} \\
&=-\frac{1}{2}\left[3^{2}-1^{2}\right]+3[3-1]+\frac{1}{2} \cdot\left[4^{2}-3^{2}\right]-3[4-3] \\
&=-\frac{1}{2}(8)+3(2)+\frac{1}{2} \cdot(7)-3(1) \\
&=-4+6+\frac{7}{2}-3=\frac{5}{2}
\end{aligned}$
& \int_{1}^{4}|\mathrm{x}-3| \mathrm{dx}=\int_{1}^{3}-(\mathrm{x}-3) \mathrm{dx}+\int_{3}^{4}(\mathrm{x}-3) \mathrm{dx} \\
&=-\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{1}^{3}+\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{3}^{4} \\
&=-\frac{1}{2}\left[3^{2}-1^{2}\right]+3[3-1]+\frac{1}{2} \cdot\left[4^{2}-3^{2}\right]-3[4-3] \\
&=-\frac{1}{2}(8)+3(2)+\frac{1}{2} \cdot(7)-3(1) \\
&=-4+6+\frac{7}{2}-3=\frac{5}{2}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.