1.* Question. ( 2.0 pts) Considere a funçäo g definida por
g(x) \int_{0}^{x^{2}} t \cos (t / 2) d t
(a) Mostre que g^{\prime}(x)=2 x^{3} \cos \left(\frac{x^{2}}{2}\right)
(b) Calcule g(\sqrt{\pi}).

Passo 1

O enunciado define
g(x)=\int_{0}^{x^{2}}t\cos\bigl(\tfrac{t}{2}\bigr)\,dt
e pede: (a) sua derivada g'(x) e (b) o valor de g(\sqrt{\pi}).
O conteúdo principal aqui é o Teorema Fundamental do Cálculo (parte 2) junto com regra da cadeia, e depois uma integral definida usando integração por partes.

Passo 2

Para achar g'(x), aplicamos o Teorema Fundamental do Cálculo: se
g(x)=\int_{0}^{u(x)}f(t)\,dt,
então
g'(x)=f\bigl(u(x)\bigr)\,u'(x).
Aqui u(x)=x^2 e f(t)=t\cos\bigl(\tfrac{t}{2}\bigr), logo

Passo 3

g'(x) =f\bigl(x^2\bigr)\cdot (x^2)' =x^2\cos\!\bigl(\tfrac{x^2}{2}\bigr)\cdot 2x =2x^3\cos\!\bigl(\tfrac{x^2}{2}\bigr).

Passo 4

Para calcular g(\sqrt{\pi}) precisamos avaliar
g(\sqrt{\pi}) =\int_{0}^{(\sqrt{\pi})^2}t\cos\bigl(\tfrac{t}{2}\bigr)\,dt =\int_{0}^{\pi}t\cos\bigl(\tfrac{t}{2}\bigr)\,dt.
Fazemos integração por partes em
\int t\cos\bigl(\tfrac{t}{2}\bigr)\,dt.
Escolhemos u=t, dv=\cos\bigl(\tfrac{t}{2}\bigr)\,dt. Então du=dt e
v=\int\cos\bigl(\tfrac{t}{2}\bigr)\,dt =2\sin\!\bigl(\tfrac{t}{2}\bigr).
Logo
\int t\cos\bigl(\tfrac{t}{2}\bigr)\,dt =2t\sin\!\bigl(\tfrac{t}{2}\bigr) -2\int\sin\!\bigl(\tfrac{t}{2}\bigr)\,dt =2t\sin\!\bigl(\tfrac{t}{2}\bigr)+4\cos\!\bigl(\tfrac{t}{2}\bigr)+C.

Passo 5

Agora avaliamos de 0 a \pi:
g(\sqrt{\pi}) =\Bigl[2t\sin\!\bigl(\tfrac{t}{2}\bigr)+4\cos\!\bigl(\tfrac{t}{2}\bigr)\Bigr]_{0}^{\pi} =\bigl(2\pi\sin(\tfrac{\pi}{2})+4\cos(\tfrac{\pi}{2})\bigr) -\bigl(0+4\cos(0)\bigr) =2\pi-4.

Resposta

(a) \;g'(x)=2x^{3}\cos\!\bigl(\tfrac{x^{2}}{2}\bigr).
(b) \;g(\sqrt{\pi})=2\pi-4.

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