What is the integral from-1 to 5 of (x^3-3x+4) ?

The integral from-1 to 5 of (x^3-3x+4) is 144

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integral from-1 to 5 of (x^3-3x+4)

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

144

Solution steps

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

\int_{- 1}^{5} x^{3} d x = 156

\int_{- 1}^{5} 3 x d x = 36

\int_{- 1}^{5} 4 d x = 24

= 156 - 36 + 24

Simplify

= 144

\int_{0}^{3} \frac{1}{( 3 x + 4 ) ^{\frac{1}{3}}} d x

\int_{0}^{8} t (8 - t) d t

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

\int_{0}^{2} e^{x} + \frac{1}{x ^{2} + 1} d x

\int_{2}^{5} - 2 x^{2} + 12 x - 17 d x

What is the integral from-1 to 5 of (x^3-3x+4) ?
The integral from-1 to 5 of (x^3-3x+4) is 144