1) The net area between the two functions is 2.
2) The net area between the two functions is 4/3.
3) The net area between the two functions is 17/6.
4) The net area between the two functions is approximately 1.218.
5) The net area between the two functions is 1/2.
How to determine the area between two functions by definite integrals
The area between the two curves is determined by definite integrals for a interval between two values of x. A general formula for the definite integral is presented below:
[tex]A = \int\limits^{b}_{a} {[f(x) - g(x)]} \, dx[/tex] (1)
Where:
- a - Lower limit
- b - Upper limit
- f(x) - "Upper" function
- g(x) - "Lower" function
Now we proceed to solve each integral:
Case I - [tex]f(x) = \sqrt{x}[/tex] and [tex]g(x) = x^{2}[/tex]
The lower and upper limits between the two functions are 0 and 1, respectively. The definite integral is described below:
[tex]A = \int\limits^1_0 {x^{0.5}} \, dx - \int\limits^1_0 {x^{2}} \, dx[/tex]
[tex]A = 2\cdot (1^{1.5}-0^{1.5})-\frac{1}{3}\cdot (1^{3}-0^{3})[/tex]
[tex]A = 2[/tex]
The net area between the two functions is 2. [tex]\blacksquare[/tex]
Case II - [tex]f(x) = -4\cdot x[/tex] and [tex]g(x) = x^{2}+3[/tex]
The lower and upper limits between the two functions are -3 and -1, respectively. The definite integral is described below:
[tex]A = - 4 \int\limits^{-1}_{-3} {x} \, dx - \int\limits^{-1}_{-3} {x^{2}} \, dx - 3 \int\limits^{-1}_{-3}\, dx[/tex]
[tex]A = -2\cdot [(-1)^{2}-(-3)^{2}]-\frac{1}{3}\cdot [(-1)^{3}-(-3)^{3}] -3\cdot [(-1)-(-3)][/tex]
[tex]A = \frac{4}{3}[/tex]
The net area between the two functions is 4/3. [tex]\blacksquare[/tex]
Case III - [tex]f(x) = x^{2}+2[/tex] and [tex]g(x) = -x[/tex]
The definite integral is described below:
[tex]A = \int\limits^{1}_{0} {x^{2}} \, dx + 2\int\limits^{1}_{0}\, dx + \int\limits^{1}_{0} {x} \, dx[/tex]
[tex]A = \frac{1}{3}\cdot (1^{3}-0^{3}) + 2\cdot (1-0) +\frac{1}{2}\cdot (1^{2}-0^{2})[/tex]
[tex]A = \frac{17}{6}[/tex]
The net area between the two functions is 17/6. [tex]\blacksquare[/tex]
Case IV - [tex]f(x) = e^{-x}[/tex] and [tex]g(x) = -x[/tex]
The definite integral is described below:
[tex]A = \int\limits^{0}_{-1} {e^{-x}} \, dx+ \int\limits^{0}_{-1} {x} \, dx[/tex]
[tex]A = -(e^{0}-e^{1}) + \frac{1}{2}\cdot [0^{2}-(-1)^{2}][/tex]
[tex]A \approx 1.218[/tex]
The net area between the two functions is approximately 1.218. [tex]\blacksquare[/tex]
Case V - [tex]f(x) = \sin 2x[/tex] and [tex]g(x) = \sin x[/tex]
This case requires a combination of definite integrals, as f(x) may be higher that g(x) in some subintervals. The combination of definite integrals is:
[tex]A = \int\limits^{\frac{\pi}{3} }_0 {\sin 2x} \, dx - \int\limits^{\frac{\pi}{3} }_{0} {\sin x} \, dx + \int\limits^{\frac{\pi}{2} }_{\frac{\pi}{3} } {\sin x} \, dx -\int\limits^{\frac{\pi}{2} }_{\frac{\pi}{3} } {\sin 2x} \, dx[/tex]
[tex]A = -\frac{1}{2}\cdot (\cos \frac{2\pi}{3}-\cos 0)+(\cos \frac{\pi}{3}-\cos 0 ) -(\cos \frac{\pi}{2}-\cos \frac{\pi}{3} )+\frac{1}{2}\cdot (\cos \pi-\cos \frac{2\pi}{3} )[/tex]
[tex]A = \frac{1}{2}[/tex]
The net area between the two functions is 1/2. [tex]\blacksquare[/tex]
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