Question

Determine f(1) se ∫ variação de 1 a x f(x)dt= xcosπx. Use o resultado do TFC: F’(x)=f(x)

Answer 1

Nos foi dado que $F(x)=\displaystyle\int_{1}^{x}f(t)dt=x\,cos(\pi x)$

Derivando F com respeito a x:

$F'(x)=\dfrac{d}{dx}\displaystyle\int_{1}^{x}f(t)dt=\dfrac{d}{dx}[x\,cos(\pi x)]$

Pela primeira parte do Teorema Fundamental do Cálculo,

$\dfrac{d}{dx}\displaystyle\int_{a}^{x}g(t)dt=g(x)$

nos pontos onde g é contínua. Então:

$F'(x)=f(x)=\dfrac{d}{dx}[x\,cos(\pi x)]\\\\\\f(x)=cos(\pi x)\dfrac{d}{dx}x+x\dfrac{d}{dx}cos(\pi x)\\\\\\f(x)=cos(\pi x)\cdot1+x[-sen(\pi x)]\dfrac{d}{dx}(\pi x)\\\\\\f(x)=cos(\pi x)-x\,sen(\pi x)\cdot\pi\\\\\boxed{\boxed{f(x)=cos(\pi x)-\pi x\cdot sen(\pi x)}}$

Encontrando $f(1)$:

$f(1)=cos(\pi\cdot1)-\pi\cdot1\cdot sen(\pi\cdot1)\\\\f(1)=cos(\pi)-\pi\,sen(\pi)\\\\f(1)=-1-\pi\cdot0\\\\\boxed{\boxed{f(1)=-1}}$