What is the integral from 0 to 4 of 1/(sqrt(2t+1)) ?

The integral from 0 to 4 of 1/(sqrt(2t+1)) is 2

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integral from 0 to 4 of 1/(sqrt(2t+1))

\int_{0}^{4} \frac{1}{2 t + 1} d t

2

Solution steps

\int_{0}^{4} \frac{1}{2 t + 1} d t

Apply u-substitution

: \int_{1}^{3} 1 d u

= \int_{1}^{3} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{1}^{3}

Simplify

= [u]_{1}^{3}

Compute the boundaries:

2

= 2

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What is the integral from 0 to 4 of 1/(sqrt(2t+1)) ?
The integral from 0 to 4 of 1/(sqrt(2t+1)) is 2