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UPSC NDA Math Model Paper 2

Question : 17 of 120

Marks: +1, -0

Let

f: R \rightarrow R

be defined by

f(x)=\begin{cases} x+2, & \text{if } x<0 \\ |x-2|, & \text{if } x\ge 0 \end{cases}

Find

\int\limits_{-2}^{3} f(x) dx

Validate

Solution:

The given function can be summarized as follows:

x	x < 0	0 ≤ x ≤ 2	2 ≤ x
f(x)	x + 2	2 - x	x - 2

Since the given function is a multi-valued function, let us separate the given definite integral into parts where the expressions of the function are different:

\int\limits_{-2}^{3} f(x) dx = \int\limits_{-2}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx

+ \int\limits_{2}^{3} f(x) dx

= \int\limits_{-2}^{0} (x+2) dx + \int\limits_{0}^{2} (2-x) dx

+ \int\limits_{2}^{3} (x-2) dx

= \left[ \frac{x^2}{2} + 2x \right]_{-2}^{0} + \left[ 2x - \frac{x^2}{2} \right]_{0}^{2}

+ \left[ \frac{x^2}{2} - 2x \right]_{2}^{3}

= [0 - (2-4)] + [4-2-0)]

+ \left[ \frac{9}{2} - 6 - (2-4) \right]

= 2 + 2 + \frac{9}{2} - 4

= 4.5

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