Question

Evaluate the definite integrals: $\int^1_0 {{(x\ e^x +sin\ \frac{\pi\ x}{4})} \, dx$

Answer 1

Answer:

Value of the given integral is

$1+(\frac{-4}{\pi})[\sqrt{2}-1]$

Step-by-step explanation:

Formula used:

Bernoulli's formula

$\int{u}\:dv=uv-u'v_1$

$\int\limits^1_0(xe^x+sin\frac{\pi\:x}{4})\:dx\\\\=\int\limits^1_0{xe^x}\:dx+\int\limits^1_0{sin\frac{\pi\:x}{4}}\:dx\\\\=I_1+I_2$

Take

u=x

u'=1

$dv=e^x\:dx\\\\\int{dv}=\int{e^x\:dx}\\\\v=e^x\\\\v_1=e^x$

By bernoulli's formula

$\int{u}\:dv=uv-u'v_1$

$I_1=\int\limits^1_0{xe^x}\:dx\\\\I_1=[xe^x-1.e^x]^1_0\\\\I_1=[1.e-e]-[0-e^0]\\\\I_1=[e-e]-[0-1]\\\\I_1=1$

$I_2=\int\limits^1_0{sin\frac{\pi\:x}{4}}\:dx\\\\I_2=\frac{1}{\frac{\pi}{4}}[-cos\frac{\pi\:x}{4}]^1_0\\\\I_2=\frac{-4}{\pi}[cos\frac{\pi\:x}{4}]^1_0\\\\I_2=\frac{-4}{\pi}[cos\frac{\pi}{4}-cos0]\\\\I_2=\frac{-4}{\pi}[\sqrt{2}-1]$

Therefore,

$\int\limits^1_0(xe^x+sin\frac{\p\:ix}{4})\:dx\\\\=I_1+I_2\\\\=1+(\frac{-4}{\pi})[\sqrt{2}-1]$