Example 1: Definite Integral Inverse Trig

$\underset{3}{\overset{7}{\int }}\frac{1}{1+x^2}\mathit{ dx}$

Step 1: Simplify and look for algebraic rewrites.

None in this example.

$\underset{3}{\overset{7}{\int }}\frac{1}{1+x^2}\mathit{ dx}$

Step 2: Identify any term(s) that include one of the six inverse trig function special cases.

Here you have 1 chunk, and it is one of the inverse trig function special cases.

$\underset{3}{\overset{7}{\int }}\frac{1}{1+x^2}\mathit{ dx}$

Step 3: Take the antiderivativeof the inversetrig function special cases using their specific Recipe.

Chunk 1: $\frac{1}{1+x^2}$

$\underset{3}{\overset{7}{\int }}\frac{1}{1+x^2}\mathit{ dx}\mathit{=}{{\mathit{tan}}^{–1}\left(x\right)|}_{x=3}^{x=7}$

Step 4 ( Definite Integral ONLY ): Evaluate the antiderivative result using the Top – Bottom method.

$\underset{3}{\overset{7}{\int }}\frac{1}{1+x^2}\mathit{ dx}={{\mathit{tan}}^{–1}\left(x\right)|}_{x=3}^{x=7}$

$=\left({\mathit{tan}}^{–1}\left(7\right)\right)–\left({\mathit{tan}}^{–1}\left(3\right)\right)\approx 5.45$

Final Result Meaning: Remember the Definite Integral will always provide you a definite value , and the Indefinite Integral provides you a family of solutions .

The Net Area between the curve $f\left(x\right)=\frac{1}{1+x^2}$ and the x-axis on the x-interval [ 3 , 7 ] is 5.45 .

Since the final result is positive, you know without even seeing the graph that there is more area above the x-axis than below it.