What is the integral from-1 to 4 of (2x-3)(x-3) ?

The integral from-1 to 4 of (2x-3)(x-3) is 125/6

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integral from-1 to 4 of (2x-3)(x-3)

\int_{- 1}^{4} (2 x - 3) (x - 3) d x

\frac{125}{6}

Decimal

20.83333 \dots

Solution steps

\int_{- 1}^{4} (2 x - 3) (x - 3) d x

Multiply by the conjugate of

2 x - 3 : \frac{4 x ^{2} - 9}{2 x + 3}

= \int_{- 1}^{4} \frac{4 x ^{2} - 9}{2 x + 3} (x - 3) d x

Expand

\frac{4 x ^{2} - 9}{2 x + 3} (x - 3) : 2 x^{2} - 9 x + 9

= \int_{- 1}^{4} 2 x^{2} - 9 x + 9 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_{- 1}^{4} 2 x^{2} d x - \int_{- 1}^{4} 9 x d x + \int_{- 1}^{4} 9 d x

\int_{- 1}^{4} 2 x^{2} d x = \frac{130}{3}

\int_{- 1}^{4} 9 x d x = \frac{135}{2}

\int_{- 1}^{4} 9 d x = 45

= \frac{130}{3} - \frac{135}{2} + 45

Simplify

\frac{130}{3} - \frac{135}{2} + 45 : \frac{125}{6}

= \frac{125}{6}

\int_{- 4}^{4} 2 x^{3} - 2 x d x

\int_{2}^{4} x ln (x) d x

\int_{- 1}^{2} \frac{2 x}{x ^{2} + 1} d x

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

\int_{0}^{1} \frac{1}{e ^{- x} + e ^{x}} d x

What is the integral from-1 to 4 of (2x-3)(x-3) ?
The integral from-1 to 4 of (2x-3)(x-3) is 125/6