The area between the curve $y = 4 + 3 x - x^{2}$ and $x -$ axis is

\frac{125}{6}

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\frac{125}{3}

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\frac{125}{9}

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None of these

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Solution

The correct option is A $\frac{125}{6}$
$y = 4 + 3 x - x^{2}$
$y = - (x^{2} - 3 x - 4)$
$y = - (x^{2} - 4 x + x - 4)$
$y = - (x (x - 4) - 1 (x - 4))$
$y = - (x + 1) (x - 4)$
$= \int_{- 1}^{4} 4 + 3 x - x^{2} d x y = {[4 x + \frac{3}{2} x^{2} - \frac{x^{3}}{3}]}_{- 1}^{4} y = (16 + 24 - \frac{4^{3}}{3} + 4 - \frac{3}{2} - \frac{1^{3}}{3}) y = (44 - \frac{65}{3} - \frac{3}{2}) y = \frac{125}{6}$

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Similar questions

The expanded form of $(3 x - 5)^{3}$ is
(a) $27 x^{3} + 135 x^{2} + 225 x - 125$
(b) $27 x^{3} + 135 x^{2} - 225 x - 125$
(c) $27 x^{3} - 135 x^{2} + 225 x - 125$
(d) none of these

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The area between the curve y=4+3x−x2 and x− axis is

The correct option is A 1256y=4+3x−x2y=−(x2−3x−4)y=−(x2−4x+x−4)y=−(x(x−4)−1(x−4))y=−(x+1)(x−4)=∫4−14+3x−x2dxy=[4x+32x2−x33]4−1y=(16+24−433+4−32−133)y=(44−653−32)y=1256

The area between the curve $y = 4 + 3 x - x^{2}$ and $x -$ axis is