What is the integral from-1 to 1 of 1/(sqrt(5-4x)) ?

The integral from-1 to 1 of 1/(sqrt(5-4x)) is 1

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integral from-1 to 1 of 1/(sqrt(5-4x))

\int_{- 1}^{1} \frac{1}{5 - 4 x} d x

1

Solution steps

\int_{- 1}^{1} \frac{1}{5 - 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

\int_{x}^{x^{2}} ln (t) d t

\int_{0}^{5} 9 - x^{2} d x

\int_{\frac{1}{4}}^{2} ln (x) d x

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

\int_{2}^{3} - 72 x^{- 3} d x

What is the integral from-1 to 1 of 1/(sqrt(5-4x)) ?
The integral from-1 to 1 of 1/(sqrt(5-4x)) is 1