1

Problem

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

2

Step

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

3

Formula

The antiderivative of sin(x) formula is \$ \int sinx dx = - cosx \$.

4

Step

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

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

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

5

Step

The trigonometric values

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

6

Step

Now substitute the values, then after simplification

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

\$ \int_0^π sin(x) dx = 1+ 1 \$

\$ \int_0^π sin(x) dx = 2 \$

7

Solution

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

8

Sumup

Please summarize Problem, Clue, Hint, Formula, Steps and Solution

Choices

9

Choice-A

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

Wrong 1

10

Choice-B

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

Wrong 0

11

Choice-C

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

Correct 2

12

Choice-D

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

Wrong 3

13

Answer

Option

C

14

Sumup

Please summarize choices