Question

evaluate the definite integral.

$$\int_{0}^{1} 5 \cos(\pi t / 2) d t$$

Explanation:

Step1: Integrate the function

We know that the integral of $\cos(ax)$ is $\frac{1}{a}\sin(ax)+C$. For the function $5\cos(\frac{\pi t}{2})$, its integral is $5\times\frac{2}{\pi}\sin(\frac{\pi t}{2})=\frac{10}{\pi}\sin(\frac{\pi t}{2})$.

Step2: Apply the fundamental theorem of calculus

The fundamental theorem of calculus states that $\int_{a}^{b}F^\prime(t)dt = F(b)-F(a)$. Here, $F(t)=\frac{10}{\pi}\sin(\frac{\pi t}{2})$, $a = 0$, and $b = 1$.
So, $F(1)-F(0)=\frac{10}{\pi}\sin(\frac{\pi\times1}{2})-\frac{10}{\pi}\sin(\frac{\pi\times0}{2})$.
Since $\sin(\frac{\pi}{2}) = 1$ and $\sin(0)=0$, we have $\frac{10}{\pi}\times1-\frac{10}{\pi}\times0$.

Answer:

$\frac{10}{\pi}$