What is the integral from 0 to 1 of (3x+1)^{sqrt(2)} ?

The integral from 0 to 1 of (3x+1)^{sqrt(2)} is (sqrt(2)*4^{sqrt(2)+1}-4^{sqrt(2)+1}-sqrt(2)+1)/3

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integral from 0 to 1 of (3x+1)^{sqrt(2)}

\int_{0}^{1} (3 x + 1)^{2} d x

\frac{2 \cdot 4 ^{2 + 1} - 4 ^{2 + 1} - 2 + 1}{3}

Decimal Notation

3.78480 \dots

Solution steps

\int_{0}^{1} (3 x + 1)^{2} d x

Apply u-substitution

: \int_{1}^{4} \frac{u ^{2}}{3} d u

= \int_{1}^{4} \frac{u ^{2}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{1}^{4} u^{2} d u

Apply the Power Rule

: [(2 - 1) u^{2 + 1}]_{1}^{4}

= \frac{1}{3} [(2 - 1) u^{2 + 1}]_{1}^{4}

Compute the boundaries:

2 \cdot 4^{2 + 1} - 4^{2 + 1} - 2 + 1

= \frac{1}{3} (2 \cdot 4^{2 + 1} - 4^{2 + 1} - 2 + 1)

Simplify

= \frac{2 \cdot 4 ^{2 + 1} - 4 ^{2 + 1} - 2 + 1}{3}

f (x) = 8 x + 1

\frac{7 s + 10}{7 s ^{2} + 26 s + 25}

f (x) = - tan (x)

y = e^{4 x^{2}}

\int_{- 2}^{3} \frac{30}{x ^{4}} d x

What is the integral from 0 to 1 of (3x+1)^{sqrt(2)} ?
The integral from 0 to 1 of (3x+1)^{sqrt(2)} is (sqrt(2)*4^{sqrt(2)+1}-4^{sqrt(2)+1}-sqrt(2)+1)/3