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# Areas Under Curve

In this section , there is only one curve above x-axis . We need to find the area of that curve in given interval . Let f(x) be the curve above x-axis .

The area of f(x) between x=a to x= b is $$\int_{a}^{b} f(x) dx$$ is the formula .

If f(x) is below x-axis then we get Area of f(x) =$$\int_{a}^{b} -f(x) dx$$.

Let us observe some of the questions on this .

Q1) Find the area of the function f(x)=x+1 between x=0 to 3.

Sol) Area = $$\int_{0}^{3} f(x) dx$$

=$$\int_{0}^{3} (x+1)dx$$

=$$[\frac{x^{2}}{2} + x ]_{0}^{3}$$

= $$\frac{3^2}{2} + 3 - 0$$

= $$\frac{9}{2} + 3$$

=$$\frac{15}{2}$$.

Therefore the area of the function f(x) is 15/2 =7.5 sq. units.

Q2) Find the area of the curve f(x)=4-x^2 between x=-2 to x=2 .

Sol) Area = $$\int_{-2}^{2} f(x) dx$$

= $$\int_{-2}^{2} (4-x^{2}) dx$$

= $$[4x - \frac{x^{3}}{3}]_{-2}^{2}$$

=$$4(2)-\frac{(2)^{3}}{3} - 4(-2) + \frac{(-2)^{3}}{3}$$

= $$8 - \frac{8}{3} + 8 - \frac{8}{3}$$

=$$16 - \frac{16}{3}$$

=$$\frac{32}{3}$$ sq.units.

Q3) Find the area of the curve f(x)=4-x between x=-1 to  x=2.

Sol) Area = $$\int_{-1}^{2}f(x) dx$$

= $$\int_{-1}^{2} 4-x dx$$

= $$[ 4x - \frac{x^{2}}{2}]_{-1}^{2}$$

= $$4(2) - \frac{2^{2}}{2} - 4(-1) + \frac{(-1)^{2}}{2}$$

= $$8 - 2 + 4 + \frac{1}{2}$$

= $$\frac{21}{2}$$ sq. units.

Q4) Find the area of the curve f(x) = Cos x   between x=$$\frac{-\prod}{2}$$ to x= $$\frac{\prod}{2}$$.

Solution:

Area = $$\int_{\frac{-\prod}{2}}^{\frac{\prod}{2}} cos x dx$$

= $$[Sin x ]_{\frac{-\prod}{2}}^{\frac{\prod}{2}}$$

= $$Sin( \frac{\prod}{2}) - Sin ( \frac{-\prod}{2} )$$

= 1 - ( -1)

= 2

Areas Under Curve

Sample Problems Related To Areas

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