What is area f(x)=x^2+4,g(x)=-x^2-1,[0,1] ?

The answer to area f(x)=x^2+4,g(x)=-x^2-1,[0,1] is 17/3

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area f(x)=x^2+4,g(x)=-x^2-1,[0,1]

area

f (x) = x^{2} + 4, g (x) = - x^{2} - 1, [0, 1]

\frac{17}{3}

Decimal Notation

5.66666 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} x^{2} + 4 - (- x^{2} - 1) d x

Solve

\int_{0}^{1} x^{2} + 4 - (- x^{2} - 1) d x : \frac{17}{3}

The area is:

= \frac{17}{3}

f (x) = 6 x^{2} + \frac{4}{x ^{2}} - 10, at x = 1

\int 5 sin (5 x) d x

\int \frac{3 x ^{2} + x + 4}{x ^{4} + 3 x ^{2} + 2} d x

f (x) = x^{2} x^{3} + 1, at x = 2

What is area f(x)=x^2+4,g(x)=-x^2-1,[0,1] ?
The answer to area f(x)=x^2+4,g(x)=-x^2-1,[0,1] is 17/3