If integral subscript C subscript 1 end subscript F with rightwards arrow on top times d r with rightwards arrow on top equals 6 and integral subscript C subscript 2 end subscript F with rightwards arrow on top times d r with rightwards arrow on top equals 13, find A. the line integral where C is the path from P to Q over C subscript 1 and then back to P over C subscript 2. integral subscript C F with rightwards arrow on top times d r with rightwards arrow on top equals _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

To my best understanding, you're given that

where C₁ and C₂ are some arbitrary paths, evidently starting at the point P and ending at the point Q, and you have to find the value of

where C is the path consisting of C₁ (from P to Q) and the reverse of C₂ (from Q back to P). So we have

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

p ÷ r=tan(x)

hence final answer is p ÷ r=tan(x).