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Popular >

integral from-3 to-1 of (2t+5)

Solution

\int_{- 3}^{- 1} (2 t + 5) d t

Solution

2

Solution steps

\int_{- 3}^{- 1} 2 t + 5 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 3}^{- 1} 2 t d t + \int_{- 3}^{- 1} 5 d t

\int_{- 3}^{- 1} 2 t d t = - 8

\int_{- 3}^{- 1} 5 d t = 10

= - 8 + 10

Simplify

= 2

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\int_{0}^{4} x x + 1 d x

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\int_{1}^{2} 3 (2 x - 1)^{2} - 3 d x

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\int_{0}^{2} x^{3} - 4 x^{2} + 3 x d x

Frequently Asked Questions (FAQ)

What is the integral from-3 to-1 of (2t+5) ?
The integral from-3 to-1 of (2t+5) is 2