Question

Find the area of the region bounded by the parabola y = 5x2, the tangent line to this parabola at (4, 80), and the x-axis.

Answer 1

$y=5x^2\implies y'=10x$

At the point when $x=4$, the slope of the line tangent to the parabola will be $10(4)=40$. Because it passes through the point (4, 80), its equation is

$y-80=40(x-4)\implies y=40x-80$

This line intercepts the x-axis at $x=2$. If you were to plot the two functions, you'll notice that for $0<x<2$ the area can be given by integrating the vertical distance between the parabola and the x-axis, while for $2<x<4$ the area is given by integrating the vertical distance between the parabola and the tangent line. To avoid having to compute two integrals, we'll integrate along the vertical axis.

To do so, first write

$y=5x^2\implies x=\sqrt{\dfrac y5}$
$y=40x-80\implies x=\dfrac{y+80}{40}$

Now in terms of $x$, the tangent line lies "above" the parabola over the entire domain of integration, so the area would be given by the single integral

$\displaystyle\int_{y=0}^{y=80}\left(\frac{y+80}{40}-\sqrt{\dfrac y5}\right)\,\mathrm dy$
$=\left(\dfrac{y^2}{80}+2y-\dfrac{2y^{3/2}}{3\sqrt5}\right)\bigg|_{y=0}^{y=80}$
$=\dfrac{80}3$

Answer 2

The required area of the region bounded by the parabola is $\frac{80}{3}$units.

Given that,

The parabola is bounded by $y = 5x^{2}$,

And tangent of the line to parabola (4,80).

We have to find,

The area of the region bounded by the parabola.

According to the question,

Parabola, $y = 5x^{2}$

To find the tangent line,

Differentiate the equation with respect to x,

$\frac{dy}{dx} = 10x$

$y' = 10x$

At x = 4 slope of line tangent of parabola,

$y' = 10(4)\\y' = 40$

Then, the slope m = 40 .

The tangent line to this parabola at (4, 80), and the x-axis and slope m = 40.

The equation of tangent,

$y-y_1 = m(x-x_1)$

At point (4,80) at m = 40

y - 80 = 40(x - 4)

y - 80 = 40x - 160

y = 40x- 80

Total area of the region bounded by parabola is ,

$= \int\limits^2_0\5x^{2} } \, dx + \int\limits^4_2 {(5x^{2} -y)} \, dx \\= \int\limits^2_0 {5x^{2} } \, dx + \int\limits^4_2 {(5x^{2}-40x+160) } \, dx \\= [\frac{5x^{3}}{3} ]^2_0 + [\frac{5x^{3} }{3} - \frac{40x^{2} }{2} + 160x ]^4_2$

$= [\frac{5(2)^{3} }{2} - \frac{5(0)^{2} }{2} ] + [\frac{5(4)^{3} }{3} - \frac{40(4)^{2} }{2} + 160(4) - [\frac{5(2)^{3} }{3} - \frac{40(2)^{2} }{2} + 160(2) ]]$

$= \frac{40}{3} - \frac{320}{3} - 320 + 320 -\frac{240}{3} + 80 - 16\\\\= \frac{320}{3} - \frac{240}{3}\\\\= \frac{80}{3}$

Hence, The required area of the region bounded by the parabola is $\frac{80}{3}$units.

For more information about Area under Curve click the link given below.

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