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Question : 72 of 160
Marks:
+1,
-0
Solution:
I=∫e−x(x3−2x2+3x−4)‌dx
Let −x=t
−dx‌=dt
I‌=∫et(−t3−2t2−3t−4)(−dt)
‌=∫et(t3+2t2+3t+4)‌dt
‌=∫et(t3+3t2−t2−2t+5t+5−1)‌dt
‌=∫et(t3+3t2)‌dt+∫et(−t2−2t)‌dt+∫et(5t+5)‌dt−∫et‌dt
‌=ett3+et(−t2)+et(5t)−et+C
‌=et(t3−t2+5t−1)+C
‌=e−x(−x3−x2−5x−1)+C
‌=−e−x(x3+x2+5x+1)+C
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