JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 18)
Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 $$-$$ 3x2 $$-$$ 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ______________.
Answer
114
Explanation
f'(x) = 6x2 $$-$$ 6x $$-$$ 12 = 6(x $$-$$ 2) (x + 1)
Point = (2, $$-$$20) & ($$-$$1, 7)
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$$A = \int\limits_{ - 1}^0 {(2{x^3} - 3{x^2} - 12x)dx + \int\limits_0^2 {(12x + 3{x^2} - 2{x^3})\,dx} } $$
$$A = \left( {{{{x^4}} \over 2} - {x^3} - 6{x^2}} \right)_{ - 1}^0 + \left( {6{x^2} + {x^3} - {{{x^4}} \over 2}} \right)_0^2$$
4A = 114
Point = (2, $$-$$20) & ($$-$$1, 7)
$$A = \int\limits_{ - 1}^0 {(2{x^3} - 3{x^2} - 12x)dx + \int\limits_0^2 {(12x + 3{x^2} - 2{x^3})\,dx} } $$
$$A = \left( {{{{x^4}} \over 2} - {x^3} - 6{x^2}} \right)_{ - 1}^0 + \left( {6{x^2} + {x^3} - {{{x^4}} \over 2}} \right)_0^2$$
4A = 114
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