The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
Area Between Curves: $y=\sqrt{x}$ and $y=8x^2$
This problem involves finding the area enclosed by two specific curves, $y = \sqrt{x}$ and $y = 8x^2$. We need to find where they intersect and then calculate the definite integral of the difference between the upper and lower curves.
Intersection Points Identification
First, find the points where the curves intersect by setting their equations equal:
$ \sqrt{x} = 8x^2 $
Square both sides to eliminate the square root:
$ x = (8x^2)^2 $
$ x = 64x^4 $
Rearrange into a polynomial equation:
$ 64x^4 - x = 0 $
Factor out $x$:
$ x(64x^3 - 1) = 0 $
The solutions are $x=0$ and $64x^3 - 1 = 0$. Solving the second equation gives:
$ 64x^3 = 1 \implies x^3 = \frac{1}{64} \implies x = \frac{1}{4} $
The intersection points occur at $x=0$ and $x=1/4$. These serve as the limits for our integral.
Upper Function Determination
We need to determine which function is greater (the upper curve) within the interval $[0, 1/4]$. Choose a test point, like $x=1/16$:
- $y = \sqrt{1/16} = 1/4$
- $y = 8(1/16)^2 = 8(1/256) = 1/32$
Since $1/4 > 1/32$, $y = \sqrt{x}$ is the upper curve in this interval.
Integral Evaluation for Area
The area $A$ is the definite integral of the upper curve minus the lower curve from $x=0$ to $x=1/4$:
$ A = \int_{0}^{1/4} (\sqrt{x} - 8x^2) dx $
Rewrite $\sqrt{x}$ as $x^{1/2}$ and integrate:
$ A = \int_{0}^{1/4} (x^{1/2} - 8x^2) dx $
$ A = \left[ \frac{x^{3/2}}{3/2} - 8\frac{x^3}{3} \right]_{0}^{1/4} $
$ A = \left[ \frac{2}{3}x^{3/2} - \frac{8}{3}x^3 \right]_{0}^{1/4} $
Evaluate using the limits:
$ A = \left( \frac{2}{3}\left(\frac{1}{4}\right)^{3/2} - \frac{8}{3}\left(\frac{1}{4}\right)^3 \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{8}{3}(0)^3 \right) $
$ A = \left( \frac{2}{3} \cdot \frac{1}{8} - \frac{8}{3} \cdot \frac{1}{64} \right) - 0 $
$ A = \frac{2}{24} - \frac{8}{192} $
$ A = \frac{1}{12} - \frac{1}{24} $
$ A = \frac{2}{24} - \frac{1}{24} = \frac{1}{24} $
Final Area Rounding
Convert the fractional result to a decimal:
$ A = \frac{1}{24} \approx 0.041666... $
Rounding to three decimal places, the area is $0.042$. This value is consistent with the provided answer range (0.03 to 0.05).
Important Questions from Application Of Definite Integral (Area)
The equation of a closed curve in two-dimensional polar coordinates is given by $r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)$. The area enclosed by the curve is ______ (answer in integer).
Two straight lines pass through the origin $(x_0, y_0) = (0,0)$. One of them passes through the point $(x_1, y_1) = (1,3)$ and the other passes through the point $(x_2, y_2) = (1,2)$.
What is the area enclosed between the straight lines in the interval $[0, 1]$ on the x-axis?
Consider the equation for a curve, $y = f(x) = x^2 + x$.
The area enclosed by the curve, the x -axis ($y = 0$ line); the vertical lines passing through $x = 1$ and $x = 2$ is _________ (rounded off to 2 decimal places)- The area of the region (rounded off to one decimal place) enclosed between the curves $y = x$ and $y = 3\sqrt{x}$ and between the lines $x = 0$ and $x = 1$ is ________ units.
- Let $S_1$ be the plane figure consisting of the points $(x, y)$ given by the inequalities $|x-1| \le 2$ and$|y + 2| \le 3$. Let $S_2$ be the plane figure given by the inequalities $x - y \ge -2$, $y \ge 1$, and $x \le 3$.Let $S$ be the union of $S_1$ and $S_2$. The area of $S$ is
