Definite Intergrals applied to area

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1. y=sec[tex]^{2}x[/tex] and y=e[tex]^{2x}[/tex], in Quadrant I, for x[tex]\leq[/tex]1. I need to calculate the area.



2. fundamental theorem



3. I'm using 0 and 1 as my lower and upper bounds and the answer I'm getting is -1.637 which is not reasonable. When I integrate using the calculator it's coming out to be 1.557. Where am I going wrong? Here's my work: (tan(1)-e[tex]^{2}[/tex]/2) - (tan(o) - e[tex]^{0}[/tex]/2)

Thanks.
 
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Your answer is right, except that e^(2x)>sec^2(x) on that range, so you have an extra minus sign. I can't say what your calculator's problem is.
 
Alright thanks for your help. Appreciate it.