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plzz find the area under the curve of function 3^(2x+1)^0.5 from x=0 to x=4
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integral
I understand it's \[3^{\sqrt{2x+1}}\] Substitute \[2x+1 =y^2\] Differentiate both sides and you will get a equation in terms of y. Adjust the limits accordingly and integrate. I believe you will get \[2\int\limits_{?}^{?}3^{y}ydy\] Resubstitute on finding the integral.
\[\int\limits_0^4 3^{\sqrt{1+2 x}} \, dx=\frac{-24+78 \text{Log}[3]}{\text{Log}[3]^2}=51.1138 \]Plot attached.
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