$\int ( x - 6 ) ^ { 7 } d x = \frac { ( x - 6 ) ^ { 8 } } { 8 } + C$

Name: $\int ( x - 6 ) ^ { 7 } d x = \frac { ( x - 6 ) ^ { 8 } } { 8 } + C$
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Question

Quizplus · Accepted Answer

This is an application of the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for any real number $n \neq -1$. Here, $n=7$, so after integrating $(x-6)^7$, we get $\frac{(x-6)^{8}}{8} + C$.

∫(x−6)7dx=(x−6)88+C\int ( x - 6 ) ^ { 7 } d x = \frac { ( x - 6 ) ^ { 8 } } { 8 } + C∫(x−6)7dx=8(x−6)8​+C

$\int ( x - 6 ) ^ { 7 } d x = \frac { ( x - 6 ) ^ { 8 } } { 8 } + C$