Question

(1 point) Evaluate the definite integral. | << + 1)e+2+28-3 dx =

Answer 1

Let us first solve by using integration first

x²+21-3 dx (at) e 6x+1}}=4 : 6+1)?-4 8 +) e 6+43-4 - 2x +-4 2 = 22 +244-3 Now, put (+1) + . 3 264+) duz at = (x+) dx = de e Calculation as limits : * *=-33) t = 63+)* = 62)=4 KFI t = (+12= (2)=4 Here, we can observe that boath upper and lower limits for t is 4 only which meany that the definete integral will be zero For further more clarity, let us save Gruent integral - Sett det [*-*]-«[]•• X2 97-3 o dx = o +1 e

Now, further more convincingly, let us plot the graph and observe how actually the area is becoming zero.

-4 -3 13

From the graph, we can clearly observe that the curve is symmetrical about x=-1 and the area between the curve and X-axis from x=-3 to x=-1 is the same in magnitude as area between the curve and X-axis from x=-1 to x=1 but opposite in sign.

Hence the given integral resulted in ZERO