Question

Y=x2-1 y=0 x=2 x=4 аычислить площадь

Answer 1

Решение задания смотри на фотографии

Answer 2

To calculate the area between the two curves $y = x^2 - 1$ and $y = 0$ from $x = 2$ to $x = 4$, we need to find the definite integral of the positive portion of $y = x^2 - 1$ over the given interval.

The area under a curve $y = f(x)$ between two points $a$ and $b$ can be calculated using the definite integral as follows:

$\text{Area} = \int_{a}^{b} f(x) \, dx$

In this case, we want to calculate the area between $y = x^2 - 1$ and the x-axis (i.e., $y = 0$) over the interval $x = 2$ to $x = 4$:

$\text{Area} = \int_{2}^{4} (x^2 - 1) \, dx$

To find the integral, we need to perform antiderivative or integration of the function $x^2 - 1$ with respect to $x$. Let's do that step by step:

$\int (x^2 - 1) \, dx = \frac{x^3}{3} - x + C$

Now, we can find the definite integral over the interval $[2, 4]$:

$\text{Area} = \left[\frac{x^3}{3} - x\right]_{2}^{4}$ $\text{Area} = \left(\frac{4^3}{3} - 4\right) - \left(\frac{2^3}{3} - 2\right)$ $\text{Area} = \left(\frac{64}{3} - 4\right) - \left(\frac{8}{3} - 2\right)$ $\text{Area} = \frac{64}{3} - 4 - \frac{8}{3} + 2$ $\text{Area} = \frac{64 - 8}{3} - 2$ $\text{Area} = \frac{56}{3} - 2$ $\text{Area} = \frac{56 - 6}{3}$ $\text{Area} = \frac{50}{3}$

So, the area between the curves $y = x^2 - 1$ and $y = 0$ over the interval $x = 2$ to $x = 4$ is $\frac{50}{3}$ square units.