What is the integral from 4 to 5 of 2/((x-3)^3) ?

The integral from 4 to 5 of 2/((x-3)^3) is 3/4

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integral from 4 to 5 of 2/((x-3)^3)

\int_{4}^{5} \frac{2}{( x - 3 ) ^{3}} d x

\frac{3}{4}

Decimal Notation

0.75

Solution steps

\int_{4}^{5} \frac{2}{( x - 3 ) ^{3}} d x

Take the constant out:

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

= 2 \cdot \int_{4}^{5} \frac{1}{( x - 3 ) ^{3}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{u ^{3}} d u

= 2 \cdot \int_{1}^{2} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{2}

= 2 [- \frac{1}{2 u ^{2}}]_{1}^{2}

Compute the boundaries:

\frac{3}{8}

= 2 \cdot \frac{3}{8}

Simplify

= \frac{3}{4}

\int_{4}^{5} \frac{1}{3 x ^{2} - 3} d x

\int_{0}^{t} x e^{- 4 x} d x

\int_{0}^{t} \frac{3}{1 - x ^{2}} d x

\int_{- 2}^{2} x^{2} + 4 d x

\int_{2}^{5} \frac{1}{( x - 2 ) ^{1}} d x

What is the integral from 4 to 5 of 2/((x-3)^3) ?
The integral from 4 to 5 of 2/((x-3)^3) is 3/4