What is the integral from 0 to 1 of (8x^3+3e^x+7) ?

The integral from 0 to 1 of (8x^3+3e^x+7) is 3e+6

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integral from 0 to 1 of (8x^3+3e^x+7)

\int_{0}^{1} (8 x^{3} + 3 e^{x} + 7) d x

3 e + 6

Decimal

14.15484 \dots

Solution steps

\int_{0}^{1} 8 x^{3} + 3 e^{x} + 7 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} 8 x^{3} d x + \int_{0}^{1} 3 e^{x} d x + \int_{0}^{1} 7 d x

\int_{0}^{1} 8 x^{3} d x = 2

\int_{0}^{1} 3 e^{x} d x = 3 (e - 1)

\int_{0}^{1} 7 d x = 7

= 2 + 3 (e - 1) + 7

Simplify

2 + 3 (e - 1) + 7 : 3 e + 6

= 3 e + 6

\int_{- 1}^{1} 8 x - 2 d x

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\int_{1}^{8} x d x

What is the integral from 0 to 1 of (8x^3+3e^x+7) ?
The integral from 0 to 1 of (8x^3+3e^x+7) is 3e+6