What is the integral from 0 to 2 of x^6J_{3}(2x) ?

The integral from 0 to 2 of x^6J_{3}(2x) is 64J_{3}

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integral from 0 to 2 of x^6J_{3}(2x)

\int_{0}^{2} x^{6} J_{3} (2 x) d x

64 J_{3}

Solution steps

\int_{0}^{2} x^{6} J_{3} \cdot 2 x d x

Take the constant out:

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

= 2 J_{3} \cdot \int_{0}^{2} x x^{6} d x

Simplify

x x^{6} : x^{7}

= 2 J_{3} \cdot \int_{0}^{2} x^{7} d x

Apply the Power Rule

: [\frac{x ^{8}}{8}]_{0}^{2}

= 2 J_{3} [\frac{x ^{8}}{8}]_{0}^{2}

Compute the boundaries:

32

= 2 J_{3} \cdot 32

Simplify

= 64 J_{3}

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

\int_{3}^{4} \frac{1}{x - 2 x} d x

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

\int_{0}^{0.5} e^{- x^{3}} d x

\int_{- 3}^{4} ∣ x + 1 ∣ d x

What is the integral from 0 to 2 of x^6J_{3}(2x) ?
The integral from 0 to 2 of x^6J_{3}(2x) is 64J_{3}