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)

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)