Question

Find the area of the region bounded by the graphs of y=
x√
and y= -x - 1 between x=1 and x=4.

Answer 1

Question #47818, Engineering, Other

Problem

Find the area of the region bounded by the graphs of $y = \sqrt{x}$ and $y = -x - 1$ between $x = 1$ and $x = 4$ .

Solve

Plot of solution set:

We need to find a shaded area. For this we use the integral.

Let $y_{1} = \sqrt{x}$ and $y_{2} = -x - 1$ , $x_{1} = 1$ , $x_{2} = 4$

$$S = S 1 + S 2$$ $$S 1 = \int_ {x _ {1}} ^ {x _ {2}} \sqrt {x} d x = \int_ {x _ {1}} ^ {x _ {2}} x ^ {\frac {1}{2}} d x = x ^ {\frac {1}{2} + 1} \cdot \frac {1}{\frac {1}{2} + 1} \Bigg | _ {1} ^ {4} = x ^ {3 / 2} \cdot \frac {2}{3} \Bigg | _ {1} ^ {4} = 4 ^ {\frac {3}{2}} \cdot \frac {2}{3} - 1 ^ {\frac {3}{2}} \cdot \frac {2}{3} = \frac {1 4}{3}$$ $$S 2 + S 3 = \left(x _ {2} - x _ {1}\right) (0 - (- 5)) = 1 5 \quad S 3 = \frac {1}{2} \left(x _ {2} - x _ {1}\right) (- 2 - (- 5)) = \frac {9}{2}$$ $$S 2 = 1 5 - S 3 = 1 5 - \frac {9}{2} = \frac {2 1}{2} \quad S = S 1 + S 2 = \frac {1 4}{3} + \frac {2 1}{2} = \frac {2 8 + 6 3}{6} = \frac {9 1}{6}$$

Answer: area is $\frac{91}{6}$ .

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