What is the integral from 0 to 2 of 3^x(1-2^x) ?

The integral from 0 to 2 of 3^x(1-2^x) is 8/(ln(3))-(35)/(ln(6))

integral from 0 to 2 of 3^x(1-2^x)

\int_{0}^{2} 3^{x} (1 - 2^{x}) d x

\frac{8}{ln ( 3 )} - \frac{35}{ln ( 6 )}

Decimal

- 12.25195 \dots

Solution steps

\int_{0}^{2} 3^{x} (1 - 2^{x}) d x

Expand

3^{x} (1 - 2^{x}) : 3^{x} - 6^{x}

= \int_{0}^{2} 3^{x} - 6^{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}^{2} 3^{x} d x - \int_{0}^{2} 6^{x} d x

\int_{0}^{2} 3^{x} d x = \frac{8}{ln ( 3 )}

\int_{0}^{2} 6^{x} d x = \frac{35}{ln ( 6 )}

= \frac{8}{ln ( 3 )} - \frac{35}{ln ( 6 )}

\int_{- 6}^{6} ∣ x ∣ d x

\int_{1}^{6} ∣ 2 x - 4 ∣ d x

\int_{1}^{e} \frac{2 - ln ( x )}{x} d x

\int_{1}^{\infty} arctan (x) d x

\int_{1}^{2} ∣ x ∣ d x

What is the integral from 0 to 2 of 3^x(1-2^x) ?
The integral from 0 to 2 of 3^x(1-2^x) is 8/(ln(3))-(35)/(ln(6))