A calculus problem by Rohith M.Athreya
There is a common integration trick when evaluating indefinite integrals , namely, ∫ e x ( f ( x ) + f ′ ( x ) ) d x = e x f ( x ) + C .
Hence, or otherwise, calculate
∫ 0 1 e x ( 2 5 6 x 1 5 − x 1 7 − x 1 6 ) d x .
Clarification: C denotes the arbitrary constant of integration .
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2 solutions
You have missed d x at the back.
Let f ( x ) = a x 1 7 + b x 1 6 , then f ′ ( x ) = 1 7 a x 1 6 + 1 6 b x 1 5 ; and:
f ( x ) + f ′ ( x ) a x 1 7 + b x 1 6 + 1 7 a x 1 6 + 1 6 b x 1 5 a x 1 7 + ( b + 1 7 a ) x 1 6 + 1 6 b x 1 5 = 2 6 5 x 1 5 − x 1 7 − x 1 6 = 2 6 5 x 1 5 − x 1 7 − x 1 6 = 2 6 5 x 1 5 − x 1 7 − x 1 6
Equating the coefficients on both sides, we get a = − 1 and b = 1 6 . Therefore, f ( x ) = 1 6 x 1 6 − x 1 7 and we have:
∫ 0 1 e x ( 2 5 6 x 1 5 − x 1 7 − x 1 6 ) d x = e x ( 1 6 x 1 6 − x 1 7 ) ∣ ∣ ∣ ∣ 0 1 = e ( 1 6 − 1 ) = 1 5 e
∫ e x [ 2 5 6 x 1 5 − x 1 7 − x 1 6 ] d x = ∫ e x [ 2 5 6 x 1 5 + 1 6 x 1 6 − x 1 7 − 1 7 x 1 6 ] d x = − x 1 6 e x ( x − 1 6 ) + c