Question

Y=3x^2+4 ,x=-3,x=2,осьОx. Найти Sф

Answer 1

S = $\int\limits^{2}_{-3} {(3x^{2}+4-0)} \, dx = (3\frac{x^{3}}{3} + 4x) |_{-3} ^{2} = (x^{3} +4x)|_{-3}^{2} = 2^{3} - (-3)^{3} + 4*2 - 4*(-3) =55$

Answer 2

To find the area under the curve of the function $y = 3x^2 + 4$ between $x = -3$ and $x = 2$, we need to calculate the definite integral of the function over this interval.

The definite integral of a function $f(x)$ from $a$ to $b$ is denoted by $\int_{a}^{b} f(x) \, dx$ and represents the signed area between the curve and the x-axis over the interval $[a, b]$.

So, to find the area under the curve between $x = -3$ and $x = 2$, we calculate:

$S = \int_{-3}^{2} (3x^2 + 4) \, dx$

Now, let's solve the integral step-by-step:

$S = \int (3x^2 + 4) \, dx = \int 3x^2 \, dx + \int 4 \, dx$

Integrate each term:

$\int 3x^2 \, dx = x^3 + C_1$

$\int 4 \, dx = 4x + C_2$

Where $C_1$ and $C_2$ are constants of integration.

Now, evaluate the definite integral:

$S = \left[x^3 + 4x\right]_{-3}^{2} = \left[(2)^3 + 4(2)\right] - \left[(-3)^3 + 4(-3)\right]$ $S = (8 + 8) - (-27 - 12) = 16 + 39 = 55$

So, the area under the curve $y = 3x^2 + 4$ between $x = -3$ and $x = 2$ is $55$ square units.