Evaluate

∫ (x 3 + 1) dx from -2 to 2

Evaluate integral (x^3 + 1) dx from -2 to 2 - Teachoo - CBSE Class 12

Question 17 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 17 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3

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Transcript

Question 17 ∫ (x3 + 1) dx from -2 to 2 This is of form ∫ a -a f (x) dx And we now that ∫_(āˆ’š‘Ž)š‘Ž š‘“(š‘„)š‘‘š‘„=0,怗 if f(āˆ’š‘„)=āˆ’š‘“(š‘„) And ∫_(āˆ’š‘Ž)^š‘Žš‘“(š‘„)š‘‘š‘„=2∫_0^š‘Žš‘“(š‘„)š‘‘š‘„ , if f(āˆ’š‘„)=š‘“(š‘„) Now, ∫_(āˆ’2)^2(š‘„^3+1)š‘‘š‘„ = ∫_(āˆ’2)^2 š‘„^3 š‘‘š‘„ć€— + ∫_(āˆ’2)^21š‘‘š‘„ Since 怖(āˆ’š‘„)怗^3=āˆ’š‘„^3 So, ∫_(āˆ’š‘Ž)^š‘Žā–’ć€–š‘„^3 š‘‘š‘„=0,怗 And 1 is constant, so f(–x) = f(x) ∫_(āˆ’š‘Ž)^š‘Žā–’1š‘‘š‘„=2∫_0^š‘Žā–’š‘‘š‘„ = 0 + 2∫_0^2ā–’1š‘‘š‘„ = 2∫_0^2ā–’š‘‘š‘„ = 2 怖[š‘„]怗_0^2 = 2 (2 – 0) = 4

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