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

Answer on Question #50829 – Math – Integral Calculus

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

Solution

First method

To find the area of region, we have to evaluate the following double integral:

$$\begin{array}{l} \iint_{A} dx\,dy = \int_{1}^{4} dx \int_{-x-1}^{\sqrt{x}} dy = \int_{1}^{4} (\sqrt{x} + x + 1) dx = \int_{1}^{4} \sqrt{x} dx + \int_{1}^{4} x dx + \int_{1}^{4} dx = \\ = \frac{2}{3} x^{3/2} \left| \begin{array}{c} 4 \\ 1 \end{array} \right. + \frac{1}{2} x^{2} \left| \begin{array}{c} 4 \\ 1 \end{array} \right. + x \left| \begin{array}{c} 4 \\ 1 \end{array} \right. = \frac{16}{3} - \frac{2}{3} + 8 - \frac{1}{2} + 3 = \frac{91}{6} = 15.1667 \end{array}$$

where $A$ is a region bounded by $y = \sqrt{x}$ and $y = -x - 1$ between $x = 1$ and $x = 4$.

Second method

Let $S$ be the area of the given region.

We have to find the sum $\int_{1}^{4}\sqrt{x} dx + \left|\int_{1}^{4}(-x - 1)dx\right|$ . The absolute value of the definite integral in the second term is taken, because all corresponding values of integrand are negative.

$$\begin{array}{l} \int_ {1} ^ {4} \sqrt {x} d x = \frac {2}{3} x ^ {3 / 2} \Bigg | _ {1} ^ {4} = \frac {1 4}{3} \\ \int_ {1} ^ {4} (- x - 1) d x = - \frac {x ^ {2}}{2} \Bigg | _ {1} ^ {4} - x \Bigg | _ {1} ^ {4} = - 8 + \frac {1}{2} - 3 = - \frac {2 1}{2} \\ S = \frac {1 4}{3} + \left| - \frac {2 1}{2} \right| = \frac {1 4}{3} + \frac {2 1}{2} = \frac {9 1}{6} = 1 5. 1 6 6 7 \\ \end{array}$$

Answer: $\frac{91}{6}$ or 15.1667.

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