Ex 7.9, 20

Transcript

Ex7.9, 20 ∫_0^1▒(𝑥 𝑒^𝑥+sin⁡〖𝜋𝑥/4〗 ) 𝑑𝑥 Step 1 :- Let F(𝑥)=∫1▒(𝑥𝑒^𝑥+𝑠𝑖𝑛 𝜋/4 𝑥)𝑑𝑥 =∫1▒〖𝑥𝑒^𝑥 𝑑𝑥+∫1▒〖𝑠𝑖𝑛 𝜋/4 𝑥 𝑑𝑥〗〗 Solving 𝑰𝟏 ∫1▒〖𝑥𝑒^𝑥 𝑑𝑥〗 Integration by parts ∫1▒〖𝑥𝑒^𝑥 𝑑𝑥=𝑥∫1▒〖𝑒^𝑥 𝑑𝑥−∫1▒[(𝑑𝑥/𝑑𝑥) ∫1▒〖𝑒^𝑥 𝑑𝑥〗]𝑑𝑥〗〗 =𝑥𝑒^𝑥−∫1▒(+1.𝑒^𝑥 𝑑𝑥)𝑑𝑥 =𝑥𝑒^𝑥−∫1▒〖𝑒^𝑥 𝑑𝑥〗 =𝑥𝑒^𝑥−𝑒^𝑥 =𝑒^𝑥 (𝑥−1) Therefore, F(𝑥)=∫1▒〖𝑥𝑒^𝑥 𝑑𝑥+∫1▒〖𝑠𝑖𝑛 𝜋/4 𝑥 𝑑𝑥〗〗 =𝑒^𝑥 (𝑥−1)+1/(𝜋/4) (−𝑐𝑜𝑠 𝜋𝑥/4) =𝑒^𝑥 (𝑥−1)−4/𝜋 𝑐𝑜𝑠 𝜋𝑥/4 Hence, F(𝑥)=𝑒^𝑥 (𝑥−1)−4/𝜋 𝑐𝑜𝑠 𝜋𝑥/4 Step 2 :- ∫_0^1▒(𝑥𝑒^𝑥+𝑠𝑖𝑛 𝜋𝑥/4) 𝑑𝑥=𝐹(1)−𝐹(0) =𝑒^1 (1−1)−4/𝜋 𝑐𝑜𝑠 ((𝜋×1))/4−𝑒^0 (0−1)+4/𝜋 𝑐𝑜𝑠 ((𝜋×0))/4 =𝑒×0−4/𝜋 𝑐𝑜𝑠 ((𝜋))/4−1(−1)+4/𝜋 cos⁡0 =(−4)/( 𝜋) 𝑐𝑜𝑠 ((𝜋))/4+1+4/𝜋 =(−4)/( 𝜋) 1/√2+1+4/𝜋 =𝟏+𝟒/𝝅−(𝟐√𝟐)/𝝅