Find the area bounded by the following: $\displaystyle y=\frac{1}{\sqrt{3x+7}} ; x=-1, x=3,x-axis $
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Originally Posted by Punch Find the area bounded by the following: $\displaystyle y=\frac{1}{\sqrt{3x+7}} ; x=-1, x=3,x-axis $ $\displaystyle \int_{-1}^{3}\frac{dx}{\sqrt{3x+7}}\,\, ;u=3x+7 \implies du=3dx \implies \frac{1}{4}\int_{4}^{16}u^{-\frac{1}{2}}du$
Originally Posted by Punch Find the area bounded by the following: $\displaystyle y=\frac{1}{\sqrt{3x+7}} ; x=-1, x=3,x-axis $ Correct me if I am wrong, but I believe the multiplier outside of the integral should be one third
sorry emptyset.. but i dont understand the part where u wrote $\displaystyle \int_{4}^{16}u^{-\frac{1}{2}}du$ isnt it suppose to be 3 on top and -1 below?
Those are the limits for x, but you are now working with the variable u = 3x+7. Check out what the values of u are when x=-1 and x=3.
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