Question

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

Answer 1

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:

$A = \int\limits^{b}_{a} {[f(x) - g(x)]} \, dx$   (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 - $f(x) = \sqrt{x}$ and $g(x) = x^{2}$

The lower and upper limits between the two functions are 0 and 1, respectively. The definite integral is described below:

$A = \int\limits^1_0 {x^{0.5}} \, dx - \int\limits^1_0 {x^{2}} \, dx$

$A = 2\cdot (1^{1.5}-0^{1.5})-\frac{1}{3}\cdot (1^{3}-0^{3})$

$A = 2$

The net area between the two functions is 2. $\blacksquare$

Case II - $f(x) = -4\cdot x$ and $g(x) = x^{2}+3$

The lower and upper limits between the two functions are -3 and -1, respectively. The definite integral is described below:

$A = - 4 \int\limits^{-1}_{-3} {x} \, dx - \int\limits^{-1}_{-3} {x^{2}} \, dx - 3 \int\limits^{-1}_{-3}\, dx$

$A = -2\cdot [(-1)^{2}-(-3)^{2}]-\frac{1}{3}\cdot [(-1)^{3}-(-3)^{3}] -3\cdot [(-1)-(-3)]$

$A = \frac{4}{3}$

The net area between the two functions is 4/3. $\blacksquare$

Case III - $f(x) = x^{2}+2$ and $g(x) = -x$

The definite integral is described below:

$A = \int\limits^{1}_{0} {x^{2}} \, dx + 2\int\limits^{1}_{0}\, dx + \int\limits^{1}_{0} {x} \, dx$

$A = \frac{1}{3}\cdot (1^{3}-0^{3}) + 2\cdot (1-0) +\frac{1}{2}\cdot (1^{2}-0^{2})$

$A = \frac{17}{6}$

The net area between the two functions is 17/6. $\blacksquare$

Case IV - $f(x) = e^{-x}$ and $g(x) = -x$

The definite integral is described below:

$A = \int\limits^{0}_{-1} {e^{-x}} \, dx+ \int\limits^{0}_{-1} {x} \, dx$

$A = -(e^{0}-e^{1}) + \frac{1}{2}\cdot [0^{2}-(-1)^{2}]$

$A \approx 1.218$

The net area between the two functions is approximately 1.218. $\blacksquare$

Case V - $f(x) = \sin 2x$ and $g(x) = \sin x$

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:

$A = \int\limits^{\frac{\pi}{3} }_0 {\sin 2x} \, dx - \int\limits^{\frac{\pi}{3} }_{0} {\sin x} \, dx + \int\limits^{\frac{\pi}{2} }_{\frac{\pi}{3} } {\sin x} \, dx -\int\limits^{\frac{\pi}{2} }_{\frac{\pi}{3} } {\sin 2x} \, dx$

$A = -\frac{1}{2}\cdot (\cos \frac{2\pi}{3}-\cos 0)+(\cos \frac{\pi}{3}-\cos 0 ) -(\cos \frac{\pi}{2}-\cos \frac{\pi}{3} )+\frac{1}{2}\cdot (\cos \pi-\cos \frac{2\pi}{3} )$

$A = \frac{1}{2}$

The net area between the two functions is 1/2. $\blacksquare$

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