Question Number 6205 by sanusihammed last updated on 18/Jun/16
$${Show}\:{that} \\ $$$$\int_{\mathrm{42}} ^{\mathrm{7}} \:\left(\mathrm{4}{x}\:−\:\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{dx}\:\:=\:\:\mathrm{15} \\ $$
Answered by FilupSmith last updated on 18/Jun/16
![∫_(42) ^( 7) (4x−1)^(1/3) dx=−∫_7 ^( 42) (4x−1)^(1/3) u=4x−1 ⇒ du=4dx =−(1/4)∫_(x=7) ^( x=42) u^(1/3) du =−(1/4)[(3/4)u^(4/3) ]_(x=7) ^(x=42) =−(3/(16))[(4x−1)^(4/3) ]_7 ^(42) =−(3/(16))((4(42)−1)^(4/3) −(4(7)−1)^(4/3) ) =−(3/(16))(167^(4/3) −27^(4/3) ) =−(3/(16))(167^(4/3) −3^4 ) ≠15](Q6209.png)
$$\int_{\mathrm{42}} ^{\:\mathrm{7}} \left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} {dx}=−\int_{\mathrm{7}} ^{\:\mathrm{42}} \left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${u}=\mathrm{4}{x}−\mathrm{1}\:\:\Rightarrow\:\:{du}=\mathrm{4}{dx} \\ $$$$ \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{{x}=\mathrm{7}} ^{\:{x}=\mathrm{42}} {u}^{\mathrm{1}/\mathrm{3}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{3}}{\mathrm{4}}{u}^{\mathrm{4}/\mathrm{3}} \right]_{{x}=\mathrm{7}} ^{{x}=\mathrm{42}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{16}}\left[\left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \right]_{\mathrm{7}} ^{\mathrm{42}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{16}}\left(\left(\mathrm{4}\left(\mathrm{42}\right)−\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} −\left(\mathrm{4}\left(\mathrm{7}\right)−\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{16}}\left(\mathrm{167}^{\mathrm{4}/\mathrm{3}} −\mathrm{27}^{\mathrm{4}/\mathrm{3}} \right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{16}}\left(\mathrm{167}^{\mathrm{4}/\mathrm{3}} −\mathrm{3}^{\mathrm{4}} \right) \\ $$$$\neq\mathrm{15} \\ $$
Commented by FilupSmith last updated on 18/Jun/16
$$\approx\mathrm{157}.\mathrm{247}\:\:\:\:\:\left(\mathrm{3}\:{d}.{p}.\right) \\ $$
Commented by sanusihammed last updated on 18/Jun/16
$${Thanks}\:{so}\:{much} \\ $$