Answer to Question #186600 in Calculus for Jethro

Question

Find the area using definite integration

1.) y = 3x² from x = 1 to x = 2

2.) xy = -1 from x = 1 to x = 2

3.) y = 4 - x² from x = -3 to x = 3

Answer 1

1) y = 3x² from x = 1 to x = 2

$\int_{1}^23x^2dx=3\int_{1}^2x^2dx=3[\frac{x^{2+1}}{2+1}]|_1^2=3[\frac{x^{3}}{3}]|_1^2=x^3|_1^2=2^3-1^3 =8 - 1=7$

2) xy = -1 from x = 1 to x = 2

$xy=-1\implies y = -\frac{1}{x}$

$\int_1^2-\frac{dx}{x}=-[ln|x|]|_1^2=-(ln(2)-ln(1)) =-ln(2)$

3) y = 4 - x² from x = -3 to x = 3

$\int_{-3}^3(4-x^2)dx=\int_{-3}^34dx-\int_{-3}^3x^2dx=4[x]|_{-3}^3-[\frac{x^3}{3}]|_{-3}^3=4(3-(-3))-(\frac{27-(-27)}{3})=12+12-(9+9)=24-18=6$