Question

$∫_0^1 (5xe^{2x}-tan(π/4)) dx=$

• A. $(\dfrac{5}{4}) * e^2 +\dfrac{1}{4}$
• B. $(\dfrac{5}{4}) * e^2 +\dfrac{9}{4}$
• C. $(\dfrac{3}{4}) * e^2 +\dfrac{1}{4}$
• D. $(\dfrac{1}{4}) * e^2 +\dfrac{5}{4}$
• E. $(\dfrac{5}{4}) * e^2$

Answer 1

Answer: (E)

$(\dfrac{5}{4}) * e^2$

Answer 2

The correct option is (E): $(\dfrac{5}{4}) * e^2$
Step 1:
Find the antiderivative of each term:
$∫(5xe^{2x}- tan(\dfrac{π}{4})) dx$
The antiderivative of $5xe^{2x}$ with respect to $x$ can be found using integration by parts.
Let's take $t = 5x$ and $dt = e^{2x} dx$.
Then, $dt = 5 dx$ and $u = \dfrac{1}{2}e^{2x}$.
Using the integration by parts formula: ∫u dv = uv - ∫v du
We get: $∫(5xe^{2x}) dx = (\frac{1}{2}) * 5x * e^{2x} - ∫(\frac{1}{2})e^{(2x)} * 5 dx$
$= \dfrac{5}{2}x * e^{2x}- \dfrac{5}{2}∫e^{2x} dx$
$= \dfrac{5}{2}x * e^{2x} - \dfrac{5}{2} * \dfrac{1}{2}e^{2x} + C_1$
= $\dfrac{5}{2}x * e^{2x} - \dfrac{5}{4}e^{2x} + C_1$
The antiderivative of $tan\,(\frac{π}{4})$ with respect to $x$ is simply -x since $tan\,(\frac{π}{4})$ is a constant.
_Step 2: _
Apply the limits:
$∫_0^1(5xe^{2x} - tan(π/4)) dx$
$=[\dfrac{5}{2}1e^{2*1} - \dfrac{5}{4}e^{2*1} + C_1] - [0 - C_1]$
$= [\dfrac{5}{2}*e^2 - \dfrac{5}{4}*e^2] + C1 - (- C1)$
$= (\dfrac{5}{2} - \dfrac{5}{4}) * e^{2} = \dfrac{5}{4} * e^2$
$= (\dfrac{5}{4}) * e^2$ (_Ans)