What is area y=sqrt(1+3x),3<= x<= 9 ?

The answer to area y=sqrt(1+3x),3<= x<= 9 is (112sqrt(7)-20sqrt(10))/9

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area y=sqrt(1+3x),3<= x<= 9

area

y = 1 + 3 x, 3 \leq x \leq 9

\frac{112 7 - 20 10}{9}

Decimal

25.89762 \dots

Solution steps

Apply the area formula:

\int_{3}^{9} 1 + 3 x d x

Solve

\int_{3}^{9} 1 + 3 x d x : \frac{112 7 - 20 10}{9}

The area is:

= \frac{112 7 - 20 10}{9}

y = 10 x^{2}, y = x^{2} + 12

\int e^{\frac{- 4 x}{3}} d x

x \to 0 lim (e^{x} - 1 - x)

n = 0 \sum \infty (\frac{2}{5})^{n}

\frac{d x}{d t} = \frac{t}{x e ^{t + 2 x}}

What is area y=sqrt(1+3x),3<= x<= 9 ?
The answer to area y=sqrt(1+3x),3<= x<= 9 is (112sqrt(7)-20sqrt(10))/9