What is area 7xe^{-x^2},y=0,0<= x<= 1 ?

The answer to area 7xe^{-x^2},y=0,0<= x<= 1 is (7(e-1))/(2e)

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area 7xe^{-x^2},y=0,0<= x<= 1

area

7 x e^{- x^{2}}, y = 0, 0 \leq x \leq 1

\frac{7 ( e - 1 )}{2 e}

Decimal Notation

2.21242 \dots

Solution steps

Apply the area formula:

\int_{0}^{1} 7 ∣ x ∣ e^{- x^{2}} d x

Solve

\int_{0}^{1} 7 ∣ x ∣ e^{- x^{2}} d x : \frac{7 ( e - 1 )}{2 e}

The area is:

= \frac{7 ( e - 1 )}{2 e}

ln (2 x - 1)

\frac{d y}{d x} + 4 y = e^{- x}

n = 1 \sum \infty \frac{n}{e ^{2 n}}

\int \frac{1}{x} d (x) d x

y^{^{'}} = \frac{( x + y + 1 )}{( - x + 2 y + 3 )}

What is area 7xe^{-x^2},y=0,0<= x<= 1 ?
The answer to area 7xe^{-x^2},y=0,0<= x<= 1 is (7(e-1))/(2e)