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)
Calculating Area Under the Curve $y = x^2 + x$
The problem asks for the area enclosed by the curve $y = f(x) = x^2 + x$, the x-axis ($y = 0$), and the vertical lines $x = 1$ and $x = 2$. This area can be found using definite integration.
Definite Integration Setup
The area ($A$) is calculated by integrating the function $f(x)$ with respect to $x$ from the lower limit ($x=1$) to the upper limit ($x=2$).
$ A = \int_{1}^{2} (x^2 + x) \,dx $Step-by-Step Integration and Calculation
- Find the antiderivative: Integrate the function term by term. $ \int (x^2 + x) \,dx = \frac{x^3}{3} + \frac{x^2}{2} $
- Evaluate the definite integral: Apply the fundamental theorem of calculus using the limits $x=2$ and $x=1$. $ A = \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{1}^{2} $
- Substitute the upper limit ($x=2$): $ \left( \frac{2^3}{3} + \frac{2^2}{2} \right) = \left( \frac{8}{3} + \frac{4}{2} \right) = \left( \frac{8}{3} + 2 \right) = \frac{8 + 6}{3} = \frac{14}{3} $
- Substitute the lower limit ($x=1$): $ \left( \frac{1^3}{3} + \frac{1^2}{2} \right) = \left( \frac{1}{3} + \frac{1}{2} \right) = \frac{2 + 3}{6} = \frac{5}{6} $
- Subtract the lower limit value from the upper limit value: $ A = \frac{14}{3} - \frac{5}{6} = \frac{28}{6} - \frac{5}{6} = \frac{23}{6} $
- Convert to decimal and round: Divide 23 by 6 and round to two decimal places. $ A = \frac{23}{6} \approx 3.8333... $ Rounded to two decimal places, $A \approx 3.83$.
The calculated area is approximately $3.83$, which lies between $3.7$ and $3.9$.
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).
The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
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?
- 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
