What is the integral from 0 to 4 of 2*6^{-2x} ?

The integral from 0 to 4 of 2*6^{-2x} is (1679615)/(1679616ln(6))

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integral from 0 to 4 of 2*6^{-2x}

\int_{0}^{4} 2 \cdot 6^{- 2 x} d x

\frac{1679615}{1679616 ln ( 6 )}

Decimal

0.55811 \dots

Solution steps

\int_{0}^{4} 2 \cdot 6^{- 2 x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{4} 6^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 8} - 3^{u} \cdot 2^{u - 1} d u

= 2 \cdot \int_{0}^{- 8} - 3^{u} \cdot 2^{u - 1} d u

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

= 2 (- \int_{- 8}^{0} - 3^{u} \cdot 2^{u - 1} d u)

Take the constant out:

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

= 2 (- (- \int_{- 8}^{0} 3^{u} \cdot 2^{u - 1} d u))

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= 2 (- (- \int_{- 8}^{0} 3^{u} \cdot 2^{u} \cdot 2^{- 1} d u))

Take the constant out:

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

= 2 (- (- 2^{- 1} \cdot \int_{- 8}^{0} 3^{u} \cdot 2^{u} d u))

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= 2 (- (- \frac{1}{2} \cdot \int_{- 8}^{0} 3^{u} \cdot 2^{u} d u))

Simplify

3^{u} \cdot 2^{u} : 6^{u}

= 2 (- (- \frac{1}{2} \cdot \int_{- 8}^{0} 6^{u} d u))

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 2 (- (- \frac{1}{2} [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}))

Simplify

2 (- (- \frac{1}{2} [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0})) : [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}

= [\frac{6 ^{u}}{ln ( 6 )}]_{- 8}^{0}

Compute the boundaries:

\frac{1679615}{1679616 ln ( 6 )}

= \frac{1679615}{1679616 ln ( 6 )}

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What is the integral from 0 to 4 of 2*6^{-2x} ?
The integral from 0 to 4 of 2*6^{-2x} is (1679615)/(1679616ln(6))