JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 18)
The area (in sq. units) of the region described by
A= {(x, y) $$\left| {} \right.$$y$$ \ge $$ x2 $$-$$ 5x + 4, x + y $$ \ge $$ 1, y $$ \le $$ 0} is :
A= {(x, y) $$\left| {} \right.$$y$$ \ge $$ x2 $$-$$ 5x + 4, x + y $$ \ge $$ 1, y $$ \le $$ 0} is :
$${7 \over 2}$$
$${{19} \over 6}$$
$${{13} \over 6}$$
$${{17} \over 6}$$
Explanation
_9th_April_Morning_Slot_en_18_1.png)
Required Area
= A1 + A2
= $$\left| {\int\limits_1^3 {\left( {1 - x} \right)} dx} \right| + \left| {\int\limits_3^4 {\left( {{x^2} - 5x + 4} \right)dx} } \right|$$
= $$\left| {\left[ {x - {{{x^2}} \over 2}} \right]_1^3} \right| + \left| {\left[ {{{{x^3}} \over 3} - {5 \over 2}{x^2} + 4x} \right]_3^4} \right|$$
= $$\left| {\left[ {\left( {3 - {9 \over 2}} \right) - \left( {1 - {1 \over 2}} \right)} \right]} \right| + \left| {\left[ {\left( {{{64} \over 3} - 40 + 16} \right) - \left( {9 - {{45} \over 2} + 12} \right)} \right]} \right|$$
= $$\left| {\left( {2 - 4} \right)} \right| + \left| {\left( {{{ - 8} \over 3} + {3 \over 2}} \right)} \right|$$
= 2 + $${7 \over 6}$$
= $${{19} \over 6}$$ sq. unit.
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