What is the integral from 1 to 2 of (7-6x)^2 ?

The integral from 1 to 2 of (7-6x)^2 is 7

Popular >

integral from 1 to 2 of (7-6x)^2

\int_{1}^{2} (7 - 6 x)^{2} d x

7

Solution steps

\int_{1}^{2} (7 - 6 x)^{2} d x

Apply u-substitution

: \int_{1}^{- 5} - \frac{u ^{2}}{6} d u

= \int_{1}^{- 5} - \frac{u ^{2}}{6} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 5}^{1} - \frac{u ^{2}}{6} d u

Take the constant out:

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

= - (- \frac{1}{6} \cdot \int_{- 5}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 5}^{1}

= - (- \frac{1}{6} [\frac{u ^{3}}{3}]_{- 5}^{1})

Simplify

= \frac{1}{6} [\frac{u ^{3}}{3}]_{- 5}^{1}

Compute the boundaries:

42

= \frac{1}{6} \cdot 42

Simplify

= 7

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

\int_{0}^{0.5} \frac{1}{( 1 - x ) ^{2}} d x

\int_{1}^{1.5} 1 + 36 x d x

\int_{0}^{4} x + 2 - x d x

\int_{- 1}^{4} (3 x^{4} - 5 x^{3} - x - 6) d x

What is the integral from 1 to 2 of (7-6x)^2 ?
The integral from 1 to 2 of (7-6x)^2 is 7