Double Integrals

Calculus Level 2

Evaluate the following integral

1 2 0 2 y ( x 2 + y 2 ) d x d y \large\int_{1}^{2}\int_{0}^{2y}( x^2+y^2)\ dx\ dy

14 3 \dfrac{14}{3} 210 2 \dfrac{210}{2} 15 4 \dfrac{15}{4} 35 2 \dfrac{35}{2}

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1 solution

1 2 0 2 y ( x 2 + y 2 ) d x d y \color{#3D99F6}\large \int_{1}^{2}\int_{0}^{2y}( x^2+y^2)dxdy

= 1 2 [ x 3 3 + y 2 x ] 0 2 y d y \color{#3D99F6}\large=\int_{1}^{2}\left[\dfrac{x^3}{3}+y^2x \right]_{0}^{2y}dy

= 1 2 [ ( 2 y ) 3 3 + y 2 ( 2 y ) ] d y \color{#3D99F6}\large=\int_{1}^{2}\left[\dfrac{(2y)^3}{3}+y^2(2y) \right]dy

= 1 2 [ 8 y 3 3 + 2 y 3 ] d y \color{#3D99F6}\large=\int_{1}^{2}\left[\dfrac{8y^3}{3}+2y^3\right]dy

= 14 3 1 2 y 3 d y \color{#3D99F6}\large=\dfrac{14}{3}\int_{1}^{2}y^3dy

= 14 3 [ y 4 4 ] 1 2 \color{#3D99F6}\large=\dfrac{14}{3}\left[\dfrac{y^4}{4}\right]_{1}^{2}

= 14 3 [ 2 4 4 1 4 4 ] \color{#3D99F6}\large=\dfrac{14}{3}\left[\dfrac{2^4}{4}-\dfrac{1^4}{4}\right]

= 14 3 ( 15 4 ) \color{#3D99F6}\large=\dfrac{14}{3}\left(\dfrac{15}{4}\right)

= 210 12 \color{#3D99F6}\large=\dfrac{210}{12}

= 35 2 \color{#3D99F6}\large=\dfrac{35}{2}

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