Question

Найдите первообразную функции y=x²-5, график которой проходит через точку (3;4).

Answer 1

F(x) = (x^3)/3 - (5x) + C
A(3;4) ∈ F(x)
x=3, y=4
(3^3)/3 - (5*3) + C = 4
9 - 15 + C = 4, C = 4 + 6 = 10
F(x) = (x^3)/3 - (5x) + 10

Answer 2

Finding the Antiderivative of y=x²-5 Passing Through the Point (3,4)

To find the antiderivative of the function y=x²-5, which passes through the point (3,4), we can use the given function and the point to determine the constant of integration.

The antiderivative of a function f(x) is denoted by F(x) and is given by the integral of f(x) with respect to x plus a constant of integration, C.

The given function is y=x²-5, and the point it passes through is (3,4).

The antiderivative F(x) of the function y=x²-5 can be found as follows:

1. Integrate the function y=x²-5 with respect to x to find F(x). 2. Use the given point (3,4) to determine the constant of integration, C.

Let's proceed with finding the antiderivative of y=x²-5 and determining the constant of integration using the given point.

The antiderivative F(x) of the function y=x²-5 is given by:

F(x) = (1/3)x³ - 5x + C

Where C is the constant of integration.

Now, we can use the given point (3,4) to determine the constant of integration, C.

When x=3, y=4:

4 = (1/3)(3)³ - 5(3) + C 4 = 9 - 15 + C 4 = -6 + C C = 4 + 6 C = 10

Therefore, the antiderivative F(x) of the function y=x²-5, passing through the point (3,4), is:

F(x) = (1/3)x³ - 5x + 10