# Definite Intergrals applied to area

1. Dec 18, 2007

### elitespart

1. y=sec$$^{2}x$$ and y=e$$^{2x}$$, in Quadrant I, for x$$\leq$$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$$^{2}$$/2) - (tan(o) - e$$^{0}$$/2)

Thanks.

2. Dec 18, 2007

### Dick

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.

3. Dec 18, 2007

### elitespart

Alright thanks for your help. Appreciate it.