Question

F(x) = -x²+6x-1 [2;4]​

Answer 1

Ответ:f(x)=4x-6x^2+1; A(0;2)

б)f(x) =1/x^2-10x^4+3; A(1;5)

Объяснение:

Answer 2

To find the integral of the function $f(x) = -x^2 + 6x - 1$ over the interval $[2, 4]$, you can use the definite integral formula:

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

In your case, $a = 2$ and $b = 4$, and $f(x) = -x^2 + 6x - 1$. So, the integral becomes:

$\int_{2}^{4} (-x^2 + 6x - 1) \, dx$

Now, let's find the antiderivative of $f(x)$:

$F(x) = \int (-x^2 + 6x - 1) \, dx$

To find $F(x)$, you can apply the power rule for integration:

$\int x^n \, dx = \frac{1}{n+1}x^{n+1} + C$

In this case:

1. For $-x^2$, the antiderivative is $-\frac{1}{3}x^3$.
2. For $6x$, the antiderivative is $3x^2$.
3. For the constant $-1$, the antiderivative is $-x$.

Now, we can write $F(x)$ as:

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

Next, we'll find $F(4)$ and $F(2)$ to calculate the definite integral over the interval $[2, 4]$:

$\int_{2}^{4} (-x^2 + 6x - 1) \, dx = F(4) - F(2)$

Substituting the values:

$F(4) = -\frac{1}{3}(4^3) + 3(4^2) - 4 + C$ $F(2) = -\frac{1}{3}(2^3) + 3(2^2) - 2 + C$

Now, calculate these values:

$F(4) = -\frac{1}{3}(64) + 3(16) - 4 + C$ $F(2) = -\frac{1}{3}(8) + 3(4) - 2 + C$

Now, you can subtract $F(2)$ from $F(4)$ to find the definite integral:

$\int_{2}^{4} (-x^2 + 6x - 1) \, dx = (F(4) - F(2))$

Calculate the values:

$\int_{2}^{4} (-x^2 + 6x - 1) \, dx = \left(-\frac{1}{3}(64) + 3(16) - 4 + C\right) - \left(-\frac{1}{3}(8) + 3(4) - 2 + C\right)$

Simplify:

$\int_{2}^{4} (-x^2 + 6x - 1) \, dx = \left(-\frac{192}{3} + 48 - 4 + C\right) - \left(-\frac{8}{3} + 12 - 2 + C\right)$