Estou com dúvida nessa questão, alguém poderia me ajudar!!!

primeira integral: \(x^4/4=1/4-(-1)^4/4=0\)

segunda integral: \(x^3/3+x^2=4^3/3+4^2-(1^3/3+1^2)=36\)

terceira integral:\(x^2+2x+1=x^3/3+x^2+x=0-[(-2)^3/3+(-2)^2+(-2)]=2/3\)

quarta integral: \(senx=t... cosxdx=dt... \int tdt=t^2/2=sen^2(x)/2=sen^2(pi/2)/2-sen^2(0)/2=1/2\)

1v,2v,3f,4v

alternativa 1

II) \(\int_{1}^{4} (x^2+2x)dx=(\dfrac{x^3}{3}+x^2)\biggr\rvert_{1}^{4}=(\dfrac{4^3}{3}+4^2)-(\dfrac{1^3}{3}+1^2)=(\dfrac{64}{3}+16)-(\dfrac{1}{3}+1)=36\)

III) \(\int_{-2}^{0} (x+1)^2dx=\dfrac{(x+1)^3}{3}\biggr\rvert_{-2}^{0}=\dfrac{(0+1)^3}{3}-\dfrac{(-2+1)^3}{3}=\dfrac{1}{3}-(-\dfrac{1}{3})=\dfrac{2}{3}\)

IV) Para resolver esta integral, pode-se usar a identidade:

\[\sin(2x)=2\sin(x)\cos(x)\]

Assim:

\[\int_{0}^{\pi/2} \sin(x) \cos(x)dx=\int_{0}^{\pi/2} \dfrac{1}{2}\sin(2x)=\dfrac{1}{2}\int_{0}^{\pi/2} \sin(2x)=\dfrac{1}{2} \cdot \dfrac{(-\cos(2x))}{2} \biggr\rvert_{0}^{\pi/2}=\dfrac{1}{2}\cdot(\dfrac{-\cos(\pi)}{2}-\dfrac{-\cos(0)}{2})=\dfrac{1}{2}\cdot (\dfrac{1}{2}+\dfrac{1}{2})=\dfrac{1}{2} \cdot 1= \dfrac{1}{2}\]

Observa-se assim que apenas as afirmativas I, II e IV estão corretas. Portanto, a resposta certa é a alternativa 1.