Question

Y=x^5+5x^4+5x^3+1,[-2;1]

Answer 1

Y=x^5+5x^4+5x³+1=-2^5+5*(-2)^4+5*(-2)³+1=-32+80-40=8-(наим)
y=x^5+5x^4+5x³+1=1^5+5*1^4+5*1³+1=1+5+5=11-(наиб)
Ответ:y(наим)=8, у(наиб)=11

Answer 2

To find the integral of the function $y = x^5 + 5x^4 + 5x^3 + 1$ over the interval $[-2, 1]$, we need to calculate the definite integral of the function with respect to $x$ over that interval.

The definite integral of a function $f(x)$ from $a$ to $b$ is denoted as $\int_{a}^{b} f(x) \, dx$.

In this case, we want to find:

$\int_{-2}^{1} (x^5 + 5x^4 + 5x^3 + 1) \, dx$

To find this integral, we need to find the antiderivative (also called indefinite integral) of the function $x^5 + 5x^4 + 5x^3 + 1$ first, and then evaluate it at the upper and lower bounds.

The antiderivative of each term can be calculated as follows:

$\int x^5 \, dx = \frac{1}{6}x^6 + C_1$ $\int 5x^4 \, dx = x^5 + C_2$ $\int 5x^3 \, dx = \frac{5}{4}x^4 + C_3$ $\int 1 \, dx = x + C_4$

Where $C_1, C_2, C_3,$ and $C_4$ are constants of integration.

Now, we can rewrite the integral as:

$\int_{-2}^{1} (x^5 + 5x^4 + 5x^3 + 1) \, dx = \left[\frac{1}{6}x^6 + x^5 + \frac{5}{4}x^4 + x\right]_{-2}^{1}$

Now, let's evaluate the expression at the upper and lower bounds:

$\left[\frac{1}{6}(1)^6 + (1)^5 + \frac{5}{4}(1)^4 + (1)\right] - \left[\frac{1}{6}(-2)^6 + (-2)^5 + \frac{5}{4}(-2)^4 + (-2)\right]$

$= \left[\frac{1}{6} + 1 + \frac{5}{4} + 1\right] - \left[\frac{1}{6}(-64) + (-32) + \frac{5}{4}(16) - 2\right]$

$= \left[\frac{1}{6} + 1 + \frac{5}{4} + 1\right] - \left[-\frac{64}{6} - 32 + 20 - 2\right]$

$= \left[\frac{1}{6} + 1 + \frac{5}{4} + 1\right] - \left[-\frac{32}{3} - 14\right]$