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Question Number 216925 by Engr_Jidda last updated on 24/Feb/25

Answered by mehdee7396 last updated on 24/Feb/25

$$s={\int }_0^1(4u+1)du+{\int }_0^2(4u+1)du+...+{\int }_0^{10}(4u+1)du$$$${={{\left(2u^2+u)\right]}_0^1+(2u^2+u)]}_0^2+...+(2u^2+u)]}_0^{10}$$$$=2(1^2+2^2+...+{10}^2)+(1+2+...+10)$$$$=2\times \frac{10\times 11\times 21}{6}+\frac{10\times 11}{2}$$$$=770+55=825$$$$$$

Commented by Engr_Jidda last updated on 25/Feb/25

$$thanks$$

Commented by Engr_Jidda last updated on 25/Feb/25

$$thanks$$

Commented by MathematicalUser2357 last updated on 26/Feb/25

$$thanks$$

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