Can yo solve this ones, please? in adittion, can you put answers and the process. The topic are area down the curve

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.

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:

Now we proceed to solve each integral:

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]

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]

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]

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]

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]

To learn more on definite integrals, we kindly invite to check this verified question: https://brainly.com/question/14279102

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