The area bounded by the curve y -4 x -3 and y -0 between x -2 and x -4 isA- -20B- 26C- -18D- 18

The correct option is C 18Area bounded by the curve is given by A=∫_x_1^x_2ydx ⇒ ∫_2^4( 4x+3) dx ⇒ [ 2x^2+3x]_2^4 ⇒ ( 2(4)^2+3×4)) ( 2(2^2)+3×2) ⇒ 32+12+8 ...

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Standard XII

Physics

The Problem of Areas

The area boun...

Question

The area bounded by the curve $y = - 4 x + 3$ and $y = 0$ between $x = 2$ and $x = 4$ is

18

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- 20

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- 18

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26

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Solution

The correct option is C $- 18$
Area bounded by the curve is given by $A = \int_{x_{1}}^{x_{2}} y d x$
$\Rightarrow \int_{2}^{4} (- 4 x + 3) d x$
$\Rightarrow {[- 2 x^{2} + 3 x]}_{2}^{2}$
$\Rightarrow (- 2 (4)^{2} + 3 \times 4)) - (- 2 (2^{2}) + 3 \times 2)$
$\Rightarrow - 32 + 12 + 8 - 6 = - 18$

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The area bounded by the curve y=−4x+3 and y=0 between x=2 and x=4 is

The correct option is C −18Area bounded by the curve is given by A=∫x2x1ydx ⇒ ∫42(−4x+3) dx ⇒ [−2x2+3x]42 ⇒ (−2(4)2+3×4))−(−2(22)+3×2) ⇒ −32+12+8−6=−18

The area bounded by the curve $y = - 4 x + 3$ and $y = 0$ between $x = 2$ and $x = 4$ is

The correct option is C $- 18$
Area bounded by the curve is given by $A = \int_{x_{1}}^{x_{2}} y d x$
$\Rightarrow \int_{2}^{4} (- 4 x + 3) d x$
$\Rightarrow {[- 2 x^{2} + 3 x]}_{2}^{2}$
$\Rightarrow (- 2 (4)^{2} + 3 \times 4)) - (- 2 (2^{2}) + 3 \times 2)$
$\Rightarrow - 32 + 12 + 8 - 6 = - 18$