Question

Найти интегралы. А) ∫(2cosx-3x2 – 3)dx Б) ∫(5-∛(х^2 ))/√хdx Если можно с объяснениями, потому что эта тема для меня непонятна

Answer 1

1)∫(2cosx - 3x² - 3)dx = ∫2cosxdx - ∫3x²dx - ∫3dx = 2 * sinx - 3x³/3 - 3x + C = 2sinx - x³ - 3x + C
2)∫((5 - ³√x²)/√x)dx = ∫5dx/√x - ∫(³√x²/√x) dx =
5∫x^(-1/2)dx - ∫x^(1/6)dx = 5x^(1/2)/½ - x^(7/6)/7/6 + C = 10√x - 6/7x * корень шестого степеня из (х) + С

Answer 2

А) Тут нужно интегрировать по частям...
То есть, первообразная 2cosx=2sinx, 3x^2=x^3, 3=3x! Осталось только только соединить
Б) Также по частям 5=5x, x представить в виде степени (2/3), т.е. первообразная для икса будет(3x^(5/3))/5

Answer 3

A) ∫(2cosx-3x^2-3)dx

To find the integral of the function ∫(2cosx-3x^2-3)dx, we can integrate each term separately.

The integral of 2cosx can be found using the formula for the integral of cosine function, which is ∫cos(x)dx = sin(x) + C, where C is the constant of integration. Therefore, the integral of 2cosx is 2sin(x) + C.

The integral of -3x^2 can be found using the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, where n is any real number except -1. Applying this rule, the integral of -3x^2 is (-3/3)x^3 + C = -x^3 + C.

The integral of -3 can be found using the constant rule for integration, which states that ∫c dx = cx + C, where c is a constant. Applying this rule, the integral of -3 is -3x + C.

Putting it all together, the integral of (2cosx-3x^2-3)dx is:

∫(2cosx-3x^2-3)dx = 2sin(x) - x^3 - 3x + C, where C is the constant of integration.

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