What is the integral from t to t/n of De(a(t)^b) ?

The integral from t to t/n of De(a(t)^b) is Dae((t^{b+1})/(n^{b+1)(b+1)}-(t^{b+1})/(b+1))

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integral from t to t/n of De(a(t)^b)

\int_{t}^{\frac{t}{n}} De (a (t)^{b}) d t

D a e (\frac{t ^{b + 1}}{n ^{b + 1} ( b + 1 )} - \frac{t ^{b + 1}}{b + 1})

Solution steps

\int_{t}^{\frac{t}{n}} De a t^{b} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= D a e \cdot \int_{t}^{\frac{t}{n}} t^{b} d t

Apply the Power Rule

: [\frac{t ^{b + 1}}{b + 1}]_{t}^{\frac{t}{n}}

= D a e [\frac{t ^{b + 1}}{b + 1}]_{t}^{\frac{t}{n}}

Compute the boundaries:

\frac{t ^{b + 1}}{n ^{b + 1} ( b + 1 )} - \frac{t ^{b + 1}}{b + 1}

= D a e (\frac{t ^{b + 1}}{n ^{b + 1} ( b + 1 )} - \frac{t ^{b + 1}}{b + 1})

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What is the integral from t to t/n of De(a(t)^b) ?
The integral from t to t/n of De(a(t)^b) is Dae((t^{b+1})/(n^{b+1)(b+1)}-(t^{b+1})/(b+1))