How do you evaluate the definite integral ∫(sint+cost)dt from [0,π4]?
1 Answer
Nov 6, 2016
Explanation:
We have to know that:
∫(f(x)+g(x))dx=∫f(x)dx+∫g(x)dx ∫sintdt=−cost+C ∫costdt=sint+C
Combining these and keeping the bounds of the integral, we see that:
∫π40(sint+cost)dt=[−cost+sint]π40
Evaluating at
=[−cos(π4)+sin(π4)]−[−cos0+sin0]
=(−√22+√22)−(−1+0)
=0+1
=1