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

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

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

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

1

Solution steps

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

Apply u-substitution

: \int_{9}^{1} - \frac{1}{4 u} d u

= \int_{9}^{1} - \frac{1}{4 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{9} - \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{4} \cdot \int_{1}^{9} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{9}

= - (- \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9})

Simplify

- (- \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9}) : \frac{1}{4} [2 u]_{1}^{9}

= \frac{1}{4} [2 u]_{1}^{9}

Compute the boundaries:

4

= \frac{1}{4} \cdot 4

Simplify

= 1

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