Show-that-42-7-4x-1-1-3-dx-15-

Question Number 6205 by sanusihammed last updated on 18/Jun/16

S h o w t h a t

\int_{42}^{7} {(4 x - 1)}^{\frac{1}{3}} d x = 15

Answered by FilupSmith last updated on 18/Jun/16

\int_{42}^{7} {(4 x - 1)}^{1 / 3} d x = - \int_{7}^{42} {(4 x - 1)}^{1 / 3}

u = 4 x - 1 \Rightarrow d u = 4 d x

= - \frac{1}{4} \int_{x = 7}^{x = 42} u^{1 / 3} d u

= - \frac{1}{4} {[\frac{3}{4} u^{4 / 3}]}_{x = 7}^{x = 42}

= - \frac{3}{16} {[{(4 x - 1)}^{4 / 3}]}_{7}^{42}

= - \frac{3}{16} ({(4 (42) - 1)}^{4 / 3} - {(4 (7) - 1)}^{4 / 3})

= - \frac{3}{16} (167^{4 / 3} - 27^{4 / 3})

= - \frac{3}{16} (167^{4 / 3} - 3^{4})

\neq 15

Commented by FilupSmith last updated on 18/Jun/16

\approx 157.247 (3 d . p .)

Commented by sanusihammed last updated on 18/Jun/16

T h a n k s s o m u c h