"Evaluate the following integrals. ∫01x3x2+1dx" was answered by our experts

High School
Calculus and Analysis
Calculus 2
Applications of integrals

sunshine022uv

Answered question

2021-12-12

Evaluate the following integrals.

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

Answer & Explanation

ramirezhereva

Beginner2021-12-13Added 28 answers

Step 1
Given: $\int_{0}^{1} x 3^{x^{2} + 1} d x$
for evaluating given integral we substitute
$x^{2} + 1 = t$ ...(1)
so,
at,
x=0, t=1
x=1, t=2
now differentiating equation (1) with respect to t
$2 x \frac{d x}{d t} + 0 = 1$
$2 x \frac{d x}{d t} = 1$
$x d x = \frac{d t}{2}$
Step 2
now in given integral replace $x^{2} + 1$ with t, xdx with $\frac{d t}{2}$ and change limits also
so,
$\int_{0}^{1} x \cdot 3^{x^{2} + 1} d x = \frac{1}{2} \int_{1}^{2} 3^{t} d t (∵ \int a^{x} d x = \frac{a^{x}}{\ln a} + c)$
$= \frac{1}{2} {(\frac{3^{t}}{\ln 3})}_{1}^{2}$
$= \frac{1}{2 \ln 3} (3^{2} - 3^{1})$
$= \frac{1}{2 \ln 3} (9 - 3)$
$= \frac{6}{2 \ln 3}$
$= \frac{3}{\ln 3}$
hence, given integral is equal to $\frac{3}{\ln 3}$ .

Jeremy Merritt

Beginner2021-12-14Added 31 answers

\int x \cdot 3^{x^{2} + 1} d x

= \frac{1}{2} \int 3^{u} d u

\int 3^{u} d u

= \frac{3^{u}}{\ln (3)}

\frac{1}{2} \int 3^{u} d u

= \frac{3^{u}}{2 \ln (3)}

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

= \frac{3^{x^{2} + 1}}{2 \ln (3)} + C

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?