Question
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 $3.9$.
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Important Questions from Application Of Definite Integral (Area)
- The area in the first quadrant bounded by the function $y = (8 - x)$ and the coordinate axes is ______ square units (answer in integer).
- Consider the following curve in polar coordinates.
$$ r = 2 - 2 \sin \theta $$
Which one of the following is the area enclosed by the curve for $0 \leq \theta \leq 2\pi$ ? The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 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.
- The arc length of the parametric curve: $x = \cos \theta$, $y = \sin \theta$, $z = \theta$ from $\theta = 0$ to $\theta = 2\pi$ is equal to ________ (round off to one decimal place).
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