What is the integral from-1 to 4 of (3x^2+8x-5) ?

The integral from-1 to 4 of (3x^2+8x-5) is 100

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integral from-1 to 4 of (3x^2+8x-5)

\int_{- 1}^{4} (3 x^{2} + 8 x - 5) d x

100

Solution steps

\int_{- 1}^{4} 3 x^{2} + 8 x - 5 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} 3 x^{2} d x + \int_{- 1}^{4} 8 x d x - \int_{- 1}^{4} 5 d x

\int_{- 1}^{4} 3 x^{2} d x = 65

\int_{- 1}^{4} 8 x d x = 60

\int_{- 1}^{4} 5 d x = 25

= 65 + 60 - 25

Simplify

= 100

\int 2 x^{4} + e^{2 x} + sin (3 x) d x

f (x) = \frac{4}{1 + x ^{2}}

\int x^{2} \cdot e^{2 x} d x

y^{^{'}} = 1 + x - y - x y

x \to 2 lim (\frac{sin ( 2 x )}{x})

What is the integral from-1 to 4 of (3x^2+8x-5) ?
The integral from-1 to 4 of (3x^2+8x-5) is 100