∫（－2→2）（－3x^2＋4x＋5）dx
＝[－x^3＋2x^2＋5x]（－2→2）＋C
＝（－8＋8＋10）－（8＋8－10）＋C
＝10－6＋C
＝4＋C...ans
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∫（－2→2）（－3x^2＋4x＋5）dx
＝[－x^3＋2x^2＋5x]（－2→2）＋C
＝（－8＋8＋10）－（8＋8－10）＋C
＝10－6＋C
＝4＋C...ans
Users viewing this question
are also looking at these questions 😉
