Evaluate iterated integral. ∫12∫493+5yxdxdy

Question

Evaluate iterated integral.  ∫12∫493+5yxdxdy

davonliefI · Accepted Answer

We can evaluate the double integral by integrating with respect to each variable one by one according to their order in the integration term . After integrating with respect to one variable , we have to apply its limit and then we can perform integration using next variable .
The following integration formulas can be used here ,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\int k x d x = k \frac{x^{2}}{2} + C

\int \frac{1}{\sqrt{x}} d x = \int x^{- \frac{1}{2}} d x

Consider the double integral

\int_{1}^{2} \int_{4}^{9} \frac{3 + 5 y}{\sqrt{x}} d x d y

Where

\sqrt{x}

can be written as

x^{\frac{1}{2}}

using the conversion formula of radical to exponent

\sqrt{m} {a^{n}} = a^{\frac{n}{m}}

The expression can be written as

(3 + 5 y) x^{- \frac{1}{2}}

by taking

x^{\frac{1}{2}}

to the numerator.
Thus we can write the integral as

\int_{1}^{2} \int_{4}^{9} (3 + 5 y) x^{- \frac{1}{2}} d x d y

First we have to integrate with respect to x by taking y terms outside as a constant.
Use the formula

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\int_{1}^{2} (3 + 5 y) [\int_{4}^{9} x^{- \frac{1}{2}} d x] d y = \int_{1}^{2} (3 + 5 y) [{(\frac{x^{- \frac{1}{2} + 1}}{\frac{- 1}{2} + 1})}_{4}^{9}] d y

= \int_{1}^{2} (3 + 5 y) [{(\frac{x^{\frac{1}{2}}}{\frac{1}{2}})}_{4}^{9}] d y

= \int_{1}^{2} (3 + 5 y) [2 (9^{\frac{1}{2}} - 4^{\frac{1}{2}})] d y

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Evaluate iterated integral. \int_1^2\int_4^9\frac{3+5y}{\sqrt{x}}dxdy

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