Problems from Integrals of functions defined by parts

Solve the integral $\int_{0}^{4} f (x) d x$ with $f (x) = {\begin{cases} x + e^{x} & if & 0 \leq x < 2 \\ 3 & if & 2 \leq x < 3 \\ 2 + e^{2 x} & if & 3 \leq x < 4 \end{cases}$

See development and solution

Development:

$\int_{0}^{2} (x + e^{x}) d x + \int_{2}^{3} 3 d x + \int_{3}^{4} (2 + e^{2 x}) d x =$

$= [\frac{x^{2}}{2} + e^{x}]_{0}^{2} + [3 x]_{2}^{3} + [2 x]_{3}^{4} + [\frac{1}{2} e^{2 x}]_{3}^{4} =$

$= (2 + e^{2} - 1) + (9 - 6) + (8 - 6) + (\frac{1}{2} e^{8} - \frac{1}{2} e^{6}) =$

$= 6 + e^{2} + \frac{1}{2} (e^{8} - e^{6})$

Solution:

The result is $6 + e^{2} + \frac{1}{2} (e^{8} - e^{6})$

Hide solution and development

Solve the integral $\int_{- 1}^{1} | x | d x$

See development and solution

Development:

The function $f (x) = | x |$ can be expressed as a piecewise function as follows: $| x | = {\begin{array}{rcl} - x & if & x < 0 \\ x & if & x \geq 0 \end{array}$

Let's solve the integral piecewise

$\int_{- 1}^{0} (- x) d x + \int_{0}^{1} x d x =$

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

$= 0 - (- \frac{1}{2}) + \frac{1}{2} - 0 = 1$

Solution:

The result is $1$

Hide solution and development