The integral ₁17x²-x² - 6x dx MUST be evaluated by breaking it up into a sum of three integrals: [*1x²-x² - 6x dx+ 17x²-x² - 6x dx+ 1² 17x²-x²³ - 6x| dx where a = 0 C = 6 a [₁17x²-x²³ - 6x| dx = √ 17x² - x² - 6x dx = [²17x²-x² - 6x| dx = 3 •[₁17x² - x² - 6x| dx = Thus

Q: 200)** = 700

A: 20032x+2=700

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