91Ó°ÊÓ

Find the areas of the regions enclosed by the lines and curves in Exercises \(41-50 .\) $$ y=x^{2} \quad \text { and } \quad y=-x^{2}+4 x $$

Evaluate the integrals in Exercises \(13-48\) . $$ \int t^{3}\left(1+t^{4}\right)^{3} d t $$

Evaluate the integrals in Exercises \(13-48\) . $$ \int \sqrt{\frac{x-1}{x^{5}}} d x $$

Find the areas of the regions enclosed by the lines and curves in Exercises \(41-50 .\) $$ y=7-2 x^{2} \quad \text { and } \quad y=x^{2}+4 $$

Find the areas of the regions enclosed by the lines and curves in Exercises \(41-50 .\) $$ y=x^{4}-4 x^{2}+4 \quad \text { and } \quad y=x^{2} $$

Evaluate the integrals in Exercises \(13-48\) . $$ \int x^{3} \sqrt{x^{2}+1} d x $$

Evaluate the integrals in Exercises \(13-48\) . $$ \int 3 x^{5} \sqrt{x^{3}+1} d x $$

Each of the following functions solves one of the initial value problems in Exercises \(47-50 .\) Which function solves which problem? Give brief reasons for your answers. $$ \begin{array}{ll}{\text { a. } y=\int_{1}^{x} \frac{1}{t} d t-3} & {\text { b. } y=\int_{0}^{x} \sec t d t+4} \\ {\text { c. } y=\int_{-1}^{x} \sec t d t+4} & {\text { d. } y=\int_{\pi}^{x} \frac{1}{t} d t-3}\end{array} $$ $$ y^{\prime}=\sec x, \quad y(-1)=4 $$

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 49 and 50 . $$ \begin{array}{l}{\int \frac{18 \tan ^{2} x \sec ^{2} x}{\left(2+\tan ^{3} x\right)^{2}} d x} \\ {\text { a. } u=\tan x, \text { followed by } v=u^{3}, \text { then by } w=2+v} \\ {\text { b. } u=\tan ^{3} x, \text { followed by } v=2+u} \\ {\text { c. } u=2+\tan ^{3} x}\end{array} $$

Each of the following functions solves one of the initial value problems in Exercises \(47-50 .\) Which function solves which problem? Give brief reasons for your answers. $$ \begin{array}{ll}{\text { a. } y=\int_{1}^{x} \frac{1}{t} d t-3} & {\text { b. } y=\int_{0}^{x} \sec t d t+4} \\ {\text { c. } y=\int_{-1}^{x} \sec t d t+4} & {\text { d. } y=\int_{\pi}^{x} \frac{1}{t} d t-3}\end{array} $$ $$ y^{\prime}=\sec x, \quad y(0)=4 $$