∫[2,6] (1/x^2)
by hand
just use the power rule, and note thet d/dx (-1/x) = 1/x^2
∫[2,6] (1/x^2) dx = -1/x [2,6] = (-1/6)-(-1/2) = ____
∫ 1 / x² dx = - 1 / x
6
∫ 1 / x² dx =
2
- 1 / 6 - ( - 1 / 2 ) = - 1 / 6 + 1 / 2 = - 1 / 6 + 3 / 6 = 2 / 6 = 2 ∙ 1 / 2 ∙ 3 = 1 / 3
To calculate the definite integral ∫[2,6] (1/x^2) by hand, follow these steps:
1. Start by using the power rule for integration: ∫x^(-n) dx = (x^(-n+1))/(-n+1) + C, where C is the constant of integration.
2. Apply the power rule to the integral: ∫(1/x^2) dx = x^(-2+1)/(-2+1) + C = -x^(-1)/(-1) + C = -(-1/x) + C = 1/x + C.
3. Evaluate the antiderivative at the upper and lower limits of integration to find the definite integral. Plug in the upper limit, 6, and the lower limit, 2, into the antiderivative expression:
∫[2,6] (1/x^2) = [1/x] │[2,6] = (1/6) - (1/2).
4. Simplify the result:
(1/6) - (1/2) = 1/6 - 3/6 = -2/6 = -1/3.
Therefore, the definite integral of (1/x^2) from 2 to 6 is -1/3.