Further Mathematics Assignment ∫012∞​∫03∞5​ dydϰ

​∫01∞2​dϰ∫03∞5​dy

∫012∞​ (y+c)5 3dϰ

∫01∞2​(5+c)−(3+c)dϰ

∫01∞2​2dϰ

(2x+c)2 1

(2(2)+c)−(2(1)+c)

(4+c)))()−(2+c)

4−2=2units2.

2. ∫04∞​∫013x∞​2ydydϰ

∫04∞​dx∫013x∞​2ydy

∫04∞​(22y2​+c)13x​1dx

∫04∞​(y2+c)3x1​dx

∫04∞​(3x2+c)_(12+c)dx

∫04∞​ [9x2−1] dx

[39x3​−x+c]4

[3x3−x+c]40​

[3(4)3−4]

192−4=188units2⋅

Adeshola Racheal,

SSS 3,

Rex Varsity.