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प्रश्न
Sketch the graph y = |x + 1|. Evaluate\[\int\limits_{- 4}^2 \left| x + 1 \right| dx\]. What does the value of this integral represent on the graph?
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उत्तर
We have,
y = |x + 1| intersects x = –4 and x = 2 at (–4, 3) and (2, 3) respectively.
Now,
y = |x + 1|
\[ = \begin{cases}\left( x + 1 \right)&\text{ For all }x > - 1\\ - \left( x + 1 \right)&\text{ For all }x < - 1\end{cases}\]
Integral represents the area enclosed between x = −4 and x = 2
\[A = \int_{- 4}^2 \left| y \right| d x\]
\[ = \int_{- 4}^{- 1} \left| y \right| d x + \int_{- 1}^2 \left| y \right| d x\]
\[ = \int_{- 4}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^2 \left( x + 1 \right) d x\]
\[ = - \left[ \frac{x^2}{2} + x \right]_{- 4}^{- 1} + \left[ \frac{x^2}{2} + x \right]_{- 1}^2 \]
\[ = - \left[ \frac{1}{2} - 1 - \frac{16}{2} + 4 \right] + \left[ \frac{4}{2} + 2 - \frac{1}{2} + 1 \right]\]
\[ = - \left[ 3 - \frac{15}{2} \right] + \left[ 5 - \frac{1}{2} \right]\]
\[ = - 3 + \frac{15}{2} + 5 - \frac{1}{2}\]
\[ = 9\text{ sq. units }\]
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