What is the integral from 0 to 1 of 6x^2+4x-3+4e^{4x} ?

The integral from 0 to 1 of 6x^2+4x-3+4e^{4x} is e^4

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integral from 0 to 1 of 6x^2+4x-3+4e^{4x}

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 x} d x

e^{4}

Decimal

54.59815 \dots

Solution steps

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 x} 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_{0}^{1} 6 x^{2} d x + \int_{0}^{1} 4 x d x - \int_{0}^{1} 3 d x + \int_{0}^{1} 4 e^{4 x} d x

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

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 3 d x = 3

\int_{0}^{1} 4 e^{4 x} d x = e^{4} - 1

= 2 + 2 - 3 + e^{4} - 1

Simplify

= e^{4}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{t}

(5, 2), (2, - 3)

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

y = (x^{3} - 16 x)^{14}, (4, 0)

\frac{1}{3 ( 5 - x ^{3} ) ^{5}}

What is the integral from 0 to 1 of 6x^2+4x-3+4e^{4x} ?
The integral from 0 to 1 of 6x^2+4x-3+4e^{4x} is e^4