What is the integral from 0 to 1/2 of (2x+1)e^x ?

The integral from 0 to 1/2 of (2x+1)e^x is 1

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integral from 0 to 1/2 of (2x+1)e^x

\int_{0}^{\frac{1}{2}} (2 x + 1) e^{x} d x

1

Solution steps

\int_{0}^{\frac{1}{2}} (2 x + 1) e^{x} d x

Apply Integration By Parts

: [(2 x + 1) e^{x} - \int 2 e^{x} d x]_{0}^{\frac{1}{2}}

= [(2 x + 1) e^{x} - \int 2 e^{x} d x]_{0}^{\frac{1}{2}}

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

= [(2 x + 1) e^{x} - 2 e^{x}]_{0}^{\frac{1}{2}}

Expand

(2 x + 1) e^{x} - 2 e^{x} : 2 e^{x} x - e^{x}

= [2 e^{x} x - e^{x}]_{0}^{\frac{1}{2}}

Compute the boundaries:

1

= 1

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

\int_{0}^{4} k x d x

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

\int_{3}^{8} \frac{1}{1 + x - 1} d x

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

What is the integral from 0 to 1/2 of (2x+1)e^x ?
The integral from 0 to 1/2 of (2x+1)e^x is 1