Question

Find the area of the region bounded by the graphs of y=
x√andy=−x−1betweenx=1andx=4.

Answer 1

Answer on Question #45960 – Math – Integral Calculus

Question.

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

Solution.

Fig.1. Our functions.

As we know, the area under the non-negative function is defined as integral of non-negative function. But in our case, we must calculate the sum of two integrals with two functions. Because the integral gives only the area between the curve and the x-axes.

Function $\sqrt{x}$ is non-negative for $1 \leq x \leq 4$ , function $(-x - 1)$ is negative for $1 \leq x \leq 4$ . To evaluate the area bounded by the graph of $y = -x - 1$ and $x$ -axis between $x = 1$ and $x = 4$ , we take expression $(-x - 1)$ with the opposite sign in the definite integral.

So, let calculate the area between the functions:

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

Answer.

$$A = \frac {9 1}{6}$$

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