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

The answer to area y=sqrt(1+3x),1<= x<= 9 is (112sqrt(7)-16)/9

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

area

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

\frac{112 7 - 16}{9}

Decimal Notation

31.14712 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{1}^{9} 1 + 3 x d x : \frac{112 7 - 16}{9}

The area is:

= \frac{112 7 - 16}{9}

x \to \infty lim (a^{x})

\frac{d}{d x} (5 x^{5} ln (x))

\int_{3}^{4} \frac{2 x - 3}{6} d x

\int_{2}^{3} (3 - x) d x

\int p^{5} ln (p) d p

What is area y=sqrt(1+3x),1<= x<= 9 ?
The answer to area y=sqrt(1+3x),1<= x<= 9 is (112sqrt(7)-16)/9