What is the integral from 1 to 2 of 3/(t^4) ?

The integral from 1 to 2 of 3/(t^4) is 7/8

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integral from 1 to 2 of 3/(t^4)

\int_{1}^{2} \frac{3}{t ^{4}} d t

\frac{7}{8}

Decimal Notation

0.875

Solution steps

\int_{1}^{2} \frac{3}{t ^{4}} d t

Take the constant out:

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

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

Apply the Power Rule

: [- \frac{1}{3 t ^{3}}]_{1}^{2}

= 3 [- \frac{1}{3 t ^{3}}]_{1}^{2}

Compute the boundaries:

\frac{7}{24}

= 3 \cdot \frac{7}{24}

Simplify

= \frac{7}{8}

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

\int sin (10 x) cos (3 x) d x

\frac{d}{d x} (9 x + 2)

\int \frac{x + 2 x - 1}{2 x x - 1} d x

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

What is the integral from 1 to 2 of 3/(t^4) ?
The integral from 1 to 2 of 3/(t^4) is 7/8