Question

The value of integration
limit 0 to5
(2t -- 3)dt is
0
(1) 5
(3) -10
(2) 10
(4) 5​

Answer 1

Correct question:

The value of integration ∫(2t - 3)dt [0 to 5] is

(1) 5 (2) 10 (3) -10 (4) 5​

The correct answer to this question is option (2) 10

Given:

∫(2t - 3)dt [0 to 5]

To Find:

The value of integration ∫(2t - 3)dt [0 to 5] =?

Solution:

The following integration formula have been used in this question:

• ∫adx = ax + C
• ∫xⁿ dx = $\frac{x^n +1}{n+1}$+ C

∫(2t - 3)dt = ∫2tdt - ∫3dt

∫(2t - 3)dt = 2∫tdt - 3∫dt

∫(2t - 3)dt = $2\frac{t^2}{2}$ - 3t + C

∫(2t - 3)dt = t² - 3t + C

Substituting the upper and lower limits value, we get

∫(2t - 3)dt [0 to 5] = [5² - 3×5] - [0² - 3×0]

∫(2t - 3)dt [0 to 5] = [25 - 15] - [0 - 0]

∫(2t - 3)dt [0 to 5] = 10 - 0

∫(2t - 3)dt [0 to 5] = 10

The correct answer to this question is option (2) 10

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