Double Integration - Multivariable Calculus

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Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

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Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

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Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

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Knew It

Double Integration - Multivariable Calculus

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