What is the integral from-1 to 1 of 6tsqrt(4t^2+1) ?

The integral from-1 to 1 of 6tsqrt(4t^2+1) is 0

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integral from-1 to 1 of 6tsqrt(4t^2+1)

\int_{- 1}^{1} 6 t 4 t^{2} + 1 d t

0

Solution steps

\int_{- 1}^{1} 6 t 4 t^{2} + 1 d t

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

6 t 4 t^{2} + 1 :

Odd

= 0

\int_{0}^{1} - x^{2} - x + 2 d x

\int_{1}^{2} (12 x^{3} + \frac{6}{x ^{2}}) d x

\int_{\frac{π}{4}}^{\frac{π}{3}} - sec^{2} (θ) d θ

\int_{1}^{2} \frac{1}{3} d x

\int_{1}^{e^{7}} \frac{2}{x} d x

What is the integral from-1 to 1 of 6tsqrt(4t^2+1) ?
The integral from-1 to 1 of 6tsqrt(4t^2+1) is 0