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The area (in sq. units) bounded by the parabola $y = x^2 -1$ , the tangent at the point $(2,3)$ to it and y-axis is:
Option 1)
$\frac{8}{3}$
Option 2)
$\frac{32}{3}$
Option 3)
$\frac{56}{3}$
Option 4)
$\frac{14}{3}$

Answers (1)

best_answer

Introduction of area under the curve -

The area between the curve $y= f(x),x$ axis and two ordinates at the point $x=a\, and \,x= b\left ( b>a \right )$ is given by

$A= \int_{a}^{b}f(x)dx=\int_{a}^{b}ydx$

- wherein

Area along x axis -

Let $y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x)$ be two curve then area bounded between the curves and the lines

x = a and x = b is

$\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |$

- wherein

Where $\Delta y= f_{2}\left ( x \right )-f_{1}(x)$

Tangent to the parabola $y=x^{2}-1$ at $\left ( 2,3 \right )$ is $yy_{1}=2a\left ( x+4 \right )$

Required shaded area

$=ar\left ( \bigtriangleup ABC \right )-\int_{-1}^{3}\sqrt{y+1}dy$

$=\frac{1}{2}\times8\times 2-\frac{2}{3}\left ( y+1 \right )^{\frac{3}{2}}|_{-1}^{3}$

$=\frac{8}{3}sq.unit.$

Option 1)
$\frac{8}{3}$
Option 2)

$\frac{32}{3}$

Option 3)

$\frac{56}{3}$

Option 4)

$\frac{14}{3}$

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