Question

Sketch the region enclosed by the given curves and find its area.

Answer 1

The area of the region enclosed by the curves x + y = 1, x - 3 = y, y = √x and x = 0 is 16.815 square units.

How to determine the area of a region enclosed by four functions

In this question we must determine the area generated by four functions: three linear functions and a radical function. First, we plot the four functions to determine the required combinations of definite integrals need for calculation:

$A = \int\limits^{0.382}_0 {[f(x) - g(x)]} \, dx + \int\limits^2_{0.382} {[h(x) - g(x)]} \, dx$, where f(x) = √x, g(x) = x - 3 and h(x) = - x + 1.

$A = \int\limits^{0.382}_{0} {[\sqrt{x} - x + 3]} \, dx + \int\limits^{2}_{0.382} {[- 2 \cdot x + 4]} \, dx$

$A = \int\limits^{0.382}_{0} {\sqrt{x}} \, dx - \int\limits^{0.382}_{0} {x} \, dx + 3 \int\limits^{0.382}_{0}\, dx - 2 \int\limits^{2}_{0.382} {x} \, dx + 4 \int\limits^{2}_{0.382}\, dx$

$A= 2 \cdot x^{\frac{3}{2} }|_{0}^{0.382} - \frac{x^{2}}{2}|_{0}^{0.382} + 3\cdot x |_{0.382}^{2} - x^{2} |_{0.382}^{2}+4\cdot x |_{0.382}^{2}$

$A = 2 \cdot (0.382^{\frac{3}{2} }-0^{\frac{3}{2} })-\left(\frac{1}{2} \right)\cdot (0.382^{2}-0^{2}) + 3 \cdot (2 - 0.382) - (2^{2}-0.382^{2})+4\cdot (2^{2}-0.382^{2})$

A = 16.815

The area of the region enclosed by the curves x + y = 1, x - 3 = y, y = √x and x = 0 is 16.815 square units.

Remark

The statement reports an inconsistency with at least one function and needs to be modified in order to apply definite integrals in a consistent manner. New form is shown below:

Sketch the region enclosed by the given curves and find its area: (i) x + y = 1, (ii) x - 3 = y, (iii) y = √x, (iv) x = 0.

To learn more on definite integrals: https://brainly.com/question/14279102

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