What is the integral from 1 to 4 of (4x^5-6x^2-7x+3) ?

The integral from 1 to 4 of (4x^5-6x^2-7x+3) is 5121/2

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integral from 1 to 4 of (4x^5-6x^2-7x+3)

\int_{1}^{4} (4 x^{5} - 6 x^{2} - 7 x + 3) d x

\frac{5121}{2}

Decimal Notation

2560.5

Solution steps

\int_{1}^{4} 4 x^{5} - 6 x^{2} - 7 x + 3 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} 4 x^{5} d x - \int_{1}^{4} 6 x^{2} d x - \int_{1}^{4} 7 x d x + \int_{1}^{4} 3 d x

\int_{1}^{4} 4 x^{5} d x = 2730

\int_{1}^{4} 6 x^{2} d x = 126

\int_{1}^{4} 7 x d x = \frac{105}{2}

\int_{1}^{4} 3 d x = 9

= 2730 - 126 - \frac{105}{2} + 9

Simplify

2730 - 126 - \frac{105}{2} + 9 : \frac{5121}{2}

= \frac{5121}{2}

\int_{0}^{6} 2 x x^{2} + 1 d x

\int_{0}^{3 π} e^{t} 6 d t

\int_{0}^{2} 4 x^{2} - 8 x + 5 d x

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

What is the integral from 1 to 4 of (4x^5-6x^2-7x+3) ?
The integral from 1 to 4 of (4x^5-6x^2-7x+3) is 5121/2