To determine which function has the highest area under the curve between $x = 0$ and $x = 10$, we need to calculate the definite integral of each function over this interval.
Calculating Area Under Curve for Each Function
Function 1: $y = x + 8$
The area is calculated by the definite integral:
Area$_1 = \int_{0}^{10} (x + 8) dx$
Integrating, we get:
Area$_1 = \left[ \frac{x^2}{2} + 8x \right]_{0}^{10}$
Evaluating at the limits:
Area$_1 = \left( \frac{10^2}{2} + 8(10) \right) - \left( \frac{0^2}{2} + 8(0) \right)$
Area$_1 = \left( \frac{100}{2} + 80 \right) - (0) = 50 + 80 = 130$
Function 2: $y = 3x$
The area is calculated by the definite integral:
Area$_2 = \int_{0}^{10} (3x) dx$
Integrating, we get:
Area$_2 = \left[ \frac{3x^2}{2} \right]_{0}^{10}$
Evaluating at the limits:
Area$_2 = \left( \frac{3(10^2)}{2} \right) - \left( \frac{3(0^2)}{2} \right)$
Area$_2 = \frac{3(100)}{2} - 0 = \frac{300}{2} = 150$
Function 3: $y = 2x + 1$
The area is calculated by the definite integral:
Area$_3 = \int_{0}^{10} (2x + 1) dx$
Integrating, we get:
Area$_3 = \left[ x^2 + x \right]_{0}^{10}$
Evaluating at the limits:
Area$_3 = \left( 10^2 + 10 \right) - \left( 0^2 + 0 \right)$
Area$_3 = (100 + 10) - 0 = 110$
Function 4: $y = x + 2$
The area is calculated by the definite integral:
Area$_4 = \int_{0}^{10} (x + 2) dx$
Integrating, we get:
Area$_4 = \left[ \frac{x^2}{2} + 2x \right]_{0}^{10}$
Evaluating at the limits:
Area$_4 = \left( \frac{10^2}{2} + 2(10) \right) - \left( \frac{0^2}{2} + 2(0) \right)$
Area$_4 = \left( \frac{100}{2} + 20 \right) - (0) = 50 + 20 = 70$
Comparing Calculated Areas
The calculated areas for the functions between $x=0$ and $x=10$ are:
- $y = x + 8$: Area = 130
- $y = 3x$: Area = 150
- $y = 2x + 1$: Area = 110
- $y = x + 2$: Area = 70
Conclusion
Comparing the areas, the function $y = 3x$ has the highest area under the curve (150) between $x = 0$ and $x = 10$.
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?
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.
