What is the integral from 0 to T of (x^4-8x+7) ?

The integral from 0 to T of (x^4-8x+7) is (T^5)/5-4T^2+7T

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integral from 0 to T of (x^4-8x+7)

\int_{0}^{T} (x^{4} - 8 x + 7) d x

\frac{T ^{5}}{5} - 4 T^{2} + 7 T

Solution steps

\int_{0}^{T} x^{4} - 8 x + 7 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{T} x^{4} d x - \int_{0}^{T} 8 x d x + \int_{0}^{T} 7 d x

\int_{0}^{T} x^{4} d x = \frac{T ^{5}}{5}

\int_{0}^{T} 8 x d x = 4 T^{2}

\int_{0}^{T} 7 d x = 7 T

= \frac{T ^{5}}{5} - 4 T^{2} + 7 T

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What is the integral from 0 to T of (x^4-8x+7) ?
The integral from 0 to T of (x^4-8x+7) is (T^5)/5-4T^2+7T