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

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

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

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

2

Solution steps

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

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

\int_{3}^{\infty} 2 e^{- s t} d t

\int_{0}^{1} 2 t^{3} d t

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

\int_{0}^{20} 400 - x^{2} d x

\int_{0}^{\infty} e^{- y} y^{2} d y

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