Find the definite integral: \$ \int_0^π sin(x) dx\$.

To find the definite integral of the function \$f(x) = sin(x)\$ from 0 to \$ pi\$, we can use the fundamental theorem of calculus

The antiderivative of sin(x) formula is

\$ \int sinx dx = - cosx \$

Applying the theorem, and then Evaluating the antiderivative at the upper and lower limits, we get:

⇒ \$ \int_0^π sin(x) dx = (- cosx)_0^π \$
⇒ \$ \int_0^π sin(x) dx = - ( cos(pi) - (cos(0)) ) \$

The trigonometric values

\$ cos(0) = 1, and cos(pi) = - 1 \$

Now substitute the values, then after simplification

⇒ \$ \int_0^π sin(x) dx = - ( - 1 - 1 ) \$
⇒ \$ \int_0^π sin(x) dx = - (- 2) \$
⇒ \$ \int_0^π sin(x) dx = 2 \$

Therefore, the definite integral of sin(x) from 0 to \$ pi\$ is 2.

Incorrect definite integral. Correct value of integral of sin(x) over [0,π] is 2, not 1

1

Incorrect statement: Integral equals 0 for the interval [0,π] with sin(x), implying net area under the curve is zero, which is untrue

0

The integral equals 2 because the area under the curve of sin(x) from 0 to π is 2

2

Option D, which states 3, is incorrect because the evaluation of the integral yields 2, not 3

3

Option

C

Can you summarize what you’ve understood in the above steps?