How do you evaluate the definite integral #int t sqrt( t^2+1dt)# bounded by #[0,sqrt7]#?

Question

How do you evaluate the definite integral #int t sqrt( t^2+1dt)#  bounded by #[0,sqrt7]#?

Samantha Adkison · Accepted Answer

It is #int_0^sqrt7 t*sqrt(t^2+1)dt=int_0^sqrt7 1/2*(t^2+1)'*sqrt(t^2+1)dt=
int_0^sqrt7 1/2*[(t^2+1)^(3/2)/(3/2)]'dt=
1/3*[(t^2+1)^(3/2)]_0^sqrt7=
1/3 (16 sqrt(2)-1) ~~ 7.2091#

Elijah Abram · Answer

#int_0^sqrt7 tsqrt(t^2+1)" " dt=7.209138999#

From the given #int tsqrt(t^2+1)" " dt=# bounded by #[0, sqrt7]#

#int_0^sqrt7 tsqrt(t^2+1)" " dt=1/2int_0^sqrt7 2t(t^2+1)^(1/2)" "dt#

#int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3*(t^2+1)^(3/2)# from 0 to #sqrt7#

#int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3[(sqrt7^2+1)^(3/2)-(0^2+1)^(3/2)]#

#int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3[(8)^(3/2)-(+1)^(3/2)]#

#int_0^sqrt7 tsqrt(t^2+1)" " dt=7.209138999#

God bless...I hope the explanation is useful.

Evelyn Agard · Answer

undefined To evaluate the definite integral $\int_{0}^{\sqrt{7}} t\sqrt{t^2 + 1} \, dt$, you can use the trigonometric substitution method. Let $t = \sqrt{7} \tan(\theta)$. Then $dt = \sqrt{7} \sec^2(\theta) \, d\theta$.

Substituting these into the integral and performing the substitution, you'll get:

$$ \int_{0}^{\sqrt{7}} t\sqrt{t^2 + 1} \, dt = \int_{0}^{\arctan{\sqrt{7}}} \sqrt{7} \tan(\theta) \sqrt{7 \tan^2(\theta) + 1} \cdot \sqrt{7} \sec^2(\theta) \, d\theta $$

After simplifying, this integral becomes:

$$ \int_{0}^{\arctan{\sqrt{7}}} 7 \tan^2(\theta) \sec^2(\theta) \, d\theta $$

Now, you can use trigonometric identities to simplify further and integrate.